(2x+10)(x-3)=28

Solution for (2x+10)(x-3)=28 equation:

(2x+10)(x-3)=28

We move all terms to the left:

(2x+10)(x-3)-(28)=0

We multiply parentheses ..

(+2x^2-6x+10x-30)-28=0

We get rid of parentheses

2x^2-6x+10x-30-28=0

We add all the numbers together, and all the variables

2x^2+4x-58=0

a = 2; b = 4; c = -58;
Δ = b2-4ac
Δ = 42-4·2·(-58)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{30}}{2*2}=\frac{-4-4\sqrt{30}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{30}}{2*2}=\frac{-4+4\sqrt{30}}{4}
$