(2x+4)(x+5)=71

Solution for (2x+4)(x+5)=71 equation:

(2x+4)(x+5)=71

We move all terms to the left:

(2x+4)(x+5)-(71)=0

We multiply parentheses ..

(+2x^2+10x+4x+20)-71=0

We get rid of parentheses

2x^2+10x+4x+20-71=0

We add all the numbers together, and all the variables

2x^2+14x-51=0

a = 2; b = 14; c = -51;
Δ = b2-4ac
Δ = 142-4·2·(-51)
Δ = 604
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:

$\sqrt{\Delta}=\sqrt{604}=\sqrt{4*151}=\sqrt{4}*\sqrt{151}=2\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{151}}{2*2}=\frac{-14-2\sqrt{151}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{151}}{2*2}=\frac{-14+2\sqrt{151}}{4}
$