# How do you solve using the quadratic formula 2x^2 - 2x = 1?

Apr 29, 2018

${x}_{1} = \frac{1 + \sqrt{3}}{2} \mathmr{and} {x}_{2} = \frac{1 - \sqrt{3}}{2}$

#### Explanation:

$2 {x}^{2} - 2 x = 1 | - 1$
2x^2-2x-1=0|:2
${x}^{2} - x - \frac{1}{2} = 0$
${x}_{1 , 2} = - \left(\frac{p}{2}\right) \pm \sqrt{{\left(\frac{p}{2}\right)}^{2} - q}$
${x}_{1 , 2} = - \left(- \frac{1}{2}\right) \pm \sqrt{{\left(- \frac{1}{2}\right)}^{2} - \left(- \frac{1}{2}\right)}$
${x}_{1 , 2} = \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{1}{2}} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$

Apr 29, 2018

$x = \frac{1 \pm \sqrt{3}}{2}$

#### Explanation:

Minus $1$:

$2 {x}^{2} - 2 x - 1 = 0$

Use the formula:

$\Rightarrow x = - b \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

Use the quadratic to find the values:

$2 {x}^{2} - 2 x - 1$

$a = 2$

$b = - 2$

$c = - 1$

$\therefore$ $\Rightarrow x = - \left(- 2\right) \pm \frac{\sqrt{{\left(- 2\right)}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \left(2\right)}$

$\Rightarrow x = \frac{1 \pm \sqrt{3}}{2}$

$\Rightarrow \frac{1 + \sqrt{3}}{2} , \mathmr{and} \frac{1 - \sqrt{3}}{2}$

You could change these to decimals or leave it in surd form, however, I will leave in surd form.