(2x+5)(x-5)=28

Solution for (2x+5)(x-5)=28 equation:

(2x+5)(x-5)=28

We move all terms to the left:

(2x+5)(x-5)-(28)=0

We multiply parentheses ..

(+2x^2-10x+5x-25)-28=0

We get rid of parentheses

2x^2-10x+5x-25-28=0

We add all the numbers together, and all the variables

2x^2-5x-53=0

a = 2; b = -5; c = -53;
Δ = b2-4ac
Δ = -52-4·2·(-53)
Δ = 449
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{-(-5)-\sqrt{449}}{2*2}=\frac{5-\sqrt{449}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{449}}{2*2}=\frac{5+\sqrt{449}}{4}
$