# Quadratic Formula, The 合欢app免费观看 Guide

To solve a quadratic equation like:

$$\textcolor{#2d6da3}{a}x^2+\textcolor{#5cb85c}{b}x+\textcolor{#d9534f}{c}=0$$

we can use the quadratic formula: $$x=\frac{-\textcolor{#5cb85c}{b} \pm \sqrt{\textcolor{#5cb85c}{b}^2-4\textcolor{#2d6da3}{a}\textcolor{#d9534f}{c}}}{2\textcolor{#2d6da3}{a}}$$

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Let's see how to use the quadratic formula to solve:

$$2x^2-5x-3=0$$

Our equation looks like: $$\textcolor{#2d6da3}{a}x^2+\textcolor{#5cb85c}{b}x+\textcolor{#d9534f}{c}=0$$ Let's find a, b, c:

• a is the coefficient of x^2
• b is the coefficient of x
• c is the constant term
So for our equation: $$2x^2-5x-3=0$$ $$\textcolor{#2d6da3}{2}x^2+\textcolor{#5cb85c}{-5}x+\textcolor{#d9534f}{-3}=0$$ $$\textcolor{#2d6da3}{a=2}$$ $$\textcolor{#5cb85c}{b=-5}$$ $$\textcolor{#d9534f}{c=-3}$$

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Let's plug in values a=2, b=-5, c=-3 to solve 2x^2-5x-3=0: $$x=\frac{-\textcolor{#5cb85c}{b} \pm \sqrt{\textcolor{#5cb85c}{b}^2-4\textcolor{#2d6da3}{a}\textcolor{#d9534f}{c}}}{2\textcolor{#2d6da3}{a}}$$ $$x=\frac{-\textcolor{#5cb85c}{(-5)} \pm \sqrt{\textcolor{#5cb85c}{(-5)}^2-4\textcolor{#2d6da3}{(2)}\textcolor{#d9534f}{(-3)}}}{2\textcolor{#2d6da3}{(2)}}$$ We then simplify step-by-step: $$x=\frac{5 \pm \sqrt{25-(-24)}}{4}$$ $$x=\frac{5 \pm \sqrt{49}}{4}$$ $$x=\frac{5 \pm 7}{4}$$ Separate the plus-minus symbol into two solutions: $$x=\frac{5 + 7}{4} \text{\enspace\enspace or \enspace\enspace} x=\frac{5 - 7}{4}$$ $$x=\frac{12}{4} \text{\enspace\enspace or \enspace\enspace} x=\frac{-2}{4}$$ $$x=3 \text{\enspace\enspace or \enspace\enspace} x=-\frac{1}{2}$$

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So by using the quadratic formula, we find the answer: $$x=3 \text{\enspace\enspace or \enspace\enspace} x=-\frac{1}{2}$$