# How to Solve a Quadratic Equation? (+FREE Worksheet!)

Learn how to simplify and solve a Quadratic Equation in a few simple and easy steps. ## Step by step guide to Solving a Quadratic Equation

1. Write the equation in the form of: $$ax^2+bx+c=0$$
2. Factorize the quadratic and solve for the variable.
4. Quadratic formula: $$\color{blue}{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$$

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### Solving a Quadratic Equation – Example 1:

Find the solutions of each quadratic. $$x^2+7x+10=0$$

$$x^2+7x+10=0$$

You can use the factorization method. $$x^2+7x+10=0$$

$$(x+5)(x+2)=0$$
Then: $$(x=-5)$$ and $$(x=-2)$$
You can also use quadratic formula: $$=\frac{-b±\sqrt{b^2-4ac}}{2a} , a=1,b=7$$ and $$c=10$$, then: $$x=\frac{-7±\sqrt{7^2-(4)×(1)×(10)}}{2×1}$$
$$x_{1}=\frac{-7+\sqrt{7^2-(4) × (1) × (10)}}{2 × 1}= \frac{-7+\sqrt{49-40}}{2}= \frac{-7+\sqrt{9}}{2} = \frac{-7+3}{2} =\frac {-4}{2}= -2$$ ,

$$x_{2}=\frac{-7-\sqrt{7^2-(4) × (1) × (10)}}{2 × 1}= \frac{-7-\sqrt{49-40}}{2}= \frac{-7-\sqrt{9}}{2} = \frac{-7-3}{2} =\frac {-10}{2}= -5$$

### Solving a Quadratic Equation – Example 2:

Find the solutions of each quadratic. $$x^2+4x+3=0$$

Use quadratic formula: $$=\frac{-b±\sqrt{b^2-4ac}}{2a} , a=1,b=4$$ and $$c=3$$, then: $$x=\frac{-4±\sqrt{4^2-(4) × (1) × (3)}}{2 × 1}$$

$$x_{1}=\frac{-4+\sqrt{4^2-(4) × (1) × (3)}}{2 × 1}= \frac{-4+\sqrt{16-12}}{2}= \frac{-4+\sqrt{4}}{2} = \frac{-4+2}{2} =\frac {-2}{2}= -1$$,

$$x_{2}=\frac{-4-\sqrt{4^2-(4) × (1) × (3)}}{2 × 1}= \frac{-4-\sqrt{16-12}}{2}= \frac{-4-\sqrt{4}}{2} = \frac{-4-2}{2} =\frac {-6}{2}= -3$$

### Solving a Quadratic Equation – Example 3:

Find the solutions of each quadratic. $$x^2+5x-6$$

Use quadratic formula: $$=\frac{-b±\sqrt{b^2-4ac}}{2a} , a=1,b=5$$ and $$c=-6$$ , then: $$x=\frac{-5±\sqrt{5^2-(4) × (1) × (-6)}}{2 × 1}$$ ,

$$x_{1}=\frac{-5+\sqrt{5^2-(4) × (1) × (-6)}}{2 × 1}= \frac{-5+\sqrt{25+24}}{2}= \frac{-5+\sqrt{49}}{2} = \frac{-5+7}{2} =\frac {2}{2}= 1$$,

$$x_{2}=\frac{-5-\sqrt{5^2-(4) × (1) × (-6)}}{2 × 1}= \frac{-5-\sqrt{25+24}}{2}= \frac{-5-\sqrt{49}}{2} = \frac{-5-7}{2} =\frac {-12}{2}= -6$$

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### Solving a Quadratic Equation – Example 4:

Find the solutions of each quadratic. $$x^2+6x+8$$

Use quadratic formula: $$=\frac{-b±\sqrt{b^2-4ac}}{2a} , a=1,b=6$$ and $$c=8$$, then: $$x=\frac{-6±\sqrt{6^2-(4) × (1) × (8)}}{2 × 1}$$ ,

$$x_{1}=\frac{-6+\sqrt{6^2-(4) × (1) × (8)}}{2 × 1}= \frac{-6+\sqrt{36-32}}{2}= \frac{-6+\sqrt{4}}{2} = \frac{-6+2}{2} =\frac {-4}{2}= -2$$,

$$x_{2}=\frac{-6-\sqrt{6^2-(4) × (1) × (8)}}{2 × 1}= \frac{-6-\sqrt{36-32}}{2}= \frac{-6-\sqrt{4}}{2} = \frac{-6-2}{2} =\frac {-8}{2}= -4$$

## Exercises for Solving a Quadratic Equation

### Solve each equation.

• $$\color{blue}{x^2-5x-14=0}$$
• $$\color{blue}{x^2+8x+15=0}$$
• $$\color{blue}{x^2-5x-36=0}$$
• $$\color{blue}{x^2-12x+35=0}$$
• $$\color{blue}{x^2+12x+32=0}$$
• $$\color{blue}{5x^2+27x+28=0}$$

• $$\color{blue}{x=-2,x=7}$$
• $$\color{blue}{x=-3,x=-5}$$
• $$\color{blue}{x=9,x=-4}$$
• $$\color{blue}{x=7,x=5}$$
• $$\color{blue}{x=-4,x=-8}$$
• $$\color{blue}{x=-\frac{7}{5},x=-4}$$

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