(x+3)(2x-5)=20

Solution for (x+3)(2x-5)=20 equation:

(x+3)(2x-5)=20

We move all terms to the left:

(x+3)(2x-5)-(20)=0

We multiply parentheses ..

(+2x^2-5x+6x-15)-20=0

We get rid of parentheses

2x^2-5x+6x-15-20=0

We add all the numbers together, and all the variables

2x^2+x-35=0

a = 2; b = 1; c = -35;
Δ = b2-4ac
Δ = 12-4·2·(-35)
Δ = 281
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{-(1)-\sqrt{281}}{2*2}=\frac{-1-\sqrt{281}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{281}}{2*2}=\frac{-1+\sqrt{281}}{4}
$