# How do you solve 3x^2-8x-19=(x-1)^2?

$x = 5 , - 2$

#### Explanation:

$3 {x}^{2} - 8 x - 19 = {\left(x - 1\right)}^{2}$

$3 {x}^{2} - 8 x - 19 = {x}^{2} - 2 x + 1$

$3 {x}^{2} \textcolor{red}{- {x}^{2}} - 8 x \textcolor{red}{+ 2 x} - 19 \textcolor{red}{- 1} = {x}^{2} \textcolor{red}{- {x}^{2}} - 2 x \textcolor{red}{+ 2 x} + 1 \textcolor{red}{- 1}$

$2 {x}^{2} - 6 x - 20 = 0$

Let's use the quadratic formula to solve for $x$:

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

$x = \frac{6 \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(2\right) \left(- 20\right)}}{2 \left(2\right)}$

$x = \frac{6 \pm \sqrt{36 + 160}}{4}$

$x = \frac{6 \pm \sqrt{196}}{4}$

$x = \frac{6 \pm 14}{4}$

$x = \frac{20}{4} = 5 , = - \frac{8}{4} = - 2$

And we can check this by graphing both sides of the original equation (note the points of intersection at $x = 5 , - 2$:

graph{(y-3x^2+8x+19)(y-(x-1)^2)=0 [-10, 10, -2.11, 26.78]}