(x-2)(x+5)=2x^2-19

Solution for (x-2)(x+5)=2x^2-19 equation:

(x-2)(x+5)=2x^2-19

We move all terms to the left:

(x-2)(x+5)-(2x^2-19)=0

We get rid of parentheses

-2x^2+(x-2)(x+5)+19=0

We multiply parentheses ..

-2x^2+(+x^2+5x-2x-10)+19=0

We get rid of parentheses

-2x^2+x^2+5x-2x-10+19=0

We add all the numbers together, and all the variables

-1x^2+3x+9=0

a = -1; b = 3; c = +9;
Δ = b2-4ac
Δ = 32-4·(-1)·9
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{5}}{2*-1}=\frac{-3-3\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{5}}{2*-1}=\frac{-3+3\sqrt{5}}{-2}
$