x2+6x-5=10

Solution for x2+6x-5=10 equation:

x2+6x-5=10

We move all terms to the left:

x2+6x-5-(10)=0

We add all the numbers together, and all the variables

x^2+6x-15=0

a = 1; b = 6; c = -15;
Δ = b2-4ac
Δ = 62-4·1·(-15)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{6}}{2*1}=\frac{-6-4\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{6}}{2*1}=\frac{-6+4\sqrt{6}}{2}
$