(2x+3)(x-7)=40

Solution for (2x+3)(x-7)=40 equation:

(2x+3)(x-7)=40

We move all terms to the left:

(2x+3)(x-7)-(40)=0

We multiply parentheses ..

(+2x^2-14x+3x-21)-40=0

We get rid of parentheses

2x^2-14x+3x-21-40=0

We add all the numbers together, and all the variables

2x^2-11x-61=0

a = 2; b = -11; c = -61;
Δ = b2-4ac
Δ = -112-4·2·(-61)
Δ = 609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{609}}{2*2}=\frac{11-\sqrt{609}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{609}}{2*2}=\frac{11+\sqrt{609}}{4}
$