x2+10x+5=20

Solution for x2+10x+5=20 equation:

x2+10x+5=20

We move all terms to the left:

x2+10x+5-(20)=0

We add all the numbers together, and all the variables

x^2+10x-15=0

a = 1; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·1·(-15)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{10}}{2*1}=\frac{-10-4\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{10}}{2*1}=\frac{-10+4\sqrt{10}}{2}
$