# How do you solve 4+4x=2x^2 using the quadratic formula?

May 18, 2017

$x = 1 \pm \sqrt{3} \text{ }$Exact value

$x \approx - 0.732050 \ldots . \to - 0.73$ to 2 decimal places

$x \approx \textcolor{w h i t e}{-} 2.732050 \ldots \ldots \to + 2.73$ to 2 decimal places

#### Explanation:

Subtract 4 and $4 x$ from both sides giving:

$y = 0 = 2 {x}^{2} - 4 x - 4$

Consider the standard form of $y = a {x}^{2} + b x + c$

That at $y = 0$ we have $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case: $a = 2 \text{; "b=-4"; } c = - 4$

so by substitution we have:

$x = \frac{+ 4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(2\right) \left(- 4\right)}}{2 \left(2\right)}$

$x = 1 \pm \frac{\sqrt{48}}{4}$

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Note that $\frac{\sqrt{48}}{4}$ is the same as $\sqrt{\frac{48}{4} ^ 2} = \sqrt{3}$

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$x = 1 \pm \sqrt{3} \text{ }$Exact value

$x \approx - 0.732050 \ldots . \to - 0.73$ to 2 decimal places

$x \approx \textcolor{w h i t e}{-} 2.732050 \ldots \ldots \to + 2.73$ to 2 decimal places