r^2+10r=7

Solution for r^2+10r=7 equation:

r^2+10r=7

We move all terms to the left:

r^2+10r-(7)=0

a = 1; b = 10; c = -7;
Δ = b2-4ac
Δ = 102-4·1·(-7)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8\sqrt{2}}{2*1}=\frac{-10-8\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8\sqrt{2}}{2*1}=\frac{-10+8\sqrt{2}}{2}
$