(2x+10)(x+3)=63

Solution for (2x+10)(x+3)=63 equation:

(2x+10)(x+3)=63

We move all terms to the left:

(2x+10)(x+3)-(63)=0

We multiply parentheses ..

(+2x^2+6x+10x+30)-63=0

We get rid of parentheses

2x^2+6x+10x+30-63=0

We add all the numbers together, and all the variables

2x^2+16x-33=0

a = 2; b = 16; c = -33;
Δ = b2-4ac
Δ = 162-4·2·(-33)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{130}}{2*2}=\frac{-16-2\sqrt{130}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{130}}{2*2}=\frac{-16+2\sqrt{130}}{4}
$