# How do you solve (2x - 7) ( x + 1) = 6x - 19?

Nov 23, 2016

$x = \frac{3}{2} \mathmr{and} 4$

#### Explanation:

Multiply out the Left hand side of the equation
$2 {x}^{2} + 2 x - 7 x - 7 = 2 {x}^{2} - 5 x - 7$
Therefore $2 {x}^{2} - 5 x - 7 = 6 x - 19$
Collecting like terms on the left
$2 {x}^{2} - 11 x + 12 = 0$
Factorise
$\left(2 x - 3\right) \left(x - 4\right) = 0$
$x = \frac{3}{2} \mathmr{and} x = 4$

Nov 24, 2016

$x = 4 , \frac{3}{2}$

#### Explanation:

NOTE: This is a long answer.

Solve $\left(2 x - 7\right) \left(x + 1\right) = 6 x - 19$

FOIL the left-hand side.

$\left(2 x - 7\right) \left(x + 1\right) = 2 x \cdot x + 2 x \cdot 1 + \left(- 7\right) \cdot x + \left(- 7 \cdot 1\right)$

$\left(2 x - 7\right) \left(x + 1\right) = 2 {x}^{2} + 2 x - 7 x - 7$

$\left(2 x - 7\right) \left(x + 1\right) = 2 {x}^{2} - 5 x - 7$

Put the equation back together.

$2 {x}^{2} - 5 x - 7 = 6 x - 19$

Subtract $5 x$ from both sides.

$2 {x}^{2} - 6 x - 5 x - 7 = - 19$

${x}^{2} - 11 x - 7 = - 19$

Add $19$ to both sides.

$2 {x}^{2} - 11 x - 7 + 19 = 0$

$2 {x}^{2} - 11 x + 12 = 0$

This is a quadratic equation, $a x + b x + c$, where $a = 2$, $b = - 11$, and $c = 12$.

Use the quadratic formula to solve for $x$.

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

$x = \frac{- \left(- 11\right) \pm \sqrt{- {11}^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

$x = \frac{11 \pm \sqrt{121 - 96}}{4}$

$x = \frac{11 \pm \sqrt{25}}{2}$

$x = \frac{11 \pm 5}{4}$

$x = \frac{16}{4} = 4$

$x = \frac{6}{4} = \frac{3}{2}$

$x = 4 , \frac{3}{2}$