# How do you solve 32x^2-3x-14=(2x-1)^2?

May 20, 2018

$x = \frac{5}{7} \mathmr{and} x = - \frac{3}{4}$

#### Explanation:

$32 {x}^{2} - 3 x - 14 = {\left(2 x - 1\right)}^{2} \text{ } \leftarrow$ remove the brackets

$32 {x}^{2} - 3 x - 14 = 4 {x}^{2} - 4 x + 1 \text{ } \leftarrow$ make it $= 0$

$32 {x}^{2} - 4 {x}^{2} - 3 x + 4 x - 14 - 1 = 0$

$28 {x}^{2} + x - 15 = 0 \text{ } \leftarrow$ factorise the quadratic

$\left(7 x - 5\right) \left(4 x + 3\right) = 0 \text{ } \leftarrow$ solve each factor set $= 0$

If $7 x - 5 = 0 \Rightarrow \text{ } x = \frac{5}{7}$

If $4 x + 3 = 0 \Rightarrow \text{ } x = - \frac{3}{4}$

May 20, 2018

$x = \frac{5}{7} , x = - \frac{3}{4}$

#### Explanation:

$32 {x}^{2} - 3 x - 14 = {\left(2 x - 1\right)}^{2}$ expand the right side,
$32 {x}^{2} - 3 x - 14 = 4 {x}^{2} - 4 x + 1$
$32 {x}^{2} - 4 {x}^{2} - 3 x + 4 x - 14 - 1 = 0$ shift them to the left side
$28 {x}^{2} + x - 15 = 0$

by using the equation $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 28 , b = 1 , c = - 15$

sub a,b,c into the equation$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(28\right) \left(- 15\right)}}{2 \left(28\right)}$
$x = \frac{- 1 \pm \sqrt{1 + 1680}}{56}$
$x = \frac{- 1 \pm \sqrt{1681}}{56}$
$x = \frac{- 1 \pm 41}{56}$
$x = \frac{- 1 + 41}{56}$ or $x = \frac{- 1 - 41}{56}$
$x = \frac{40}{56}$ or $x = \frac{- 42}{56}$

$x = \frac{5}{7}$ or $x = \frac{- 3}{4}$