(2x-3)(x+7)=1+11x-x^2

Solution for (2x-3)(x+7)=1+11x-x^2 equation:

(2x-3)(x+7)=1+11x-x^2

We move all terms to the left:

(2x-3)(x+7)-(1+11x-x^2)=0

We get rid of parentheses

x^2-11x+(2x-3)(x+7)-1=0

We multiply parentheses ..

x^2+(+2x^2+14x-3x-21)-11x-1=0

We get rid of parentheses

x^2+2x^2+14x-3x-11x-21-1=0

We add all the numbers together, and all the variables

3x^2-22=0

a = 3; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·3·(-22)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{66}}{2*3}=\frac{0-2\sqrt{66}}{6}
=-\frac{2\sqrt{66}}{6}
=-\frac{\sqrt{66}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{66}}{2*3}=\frac{0+2\sqrt{66}}{6}
=\frac{2\sqrt{66}}{6}
=\frac{\sqrt{66}}{3}
$