r^2+6r-7=10

Solution for r^2+6r-7=10 equation:

r^2+6r-7=10

We move all terms to the left:

r^2+6r-7-(10)=0

We add all the numbers together, and all the variables

r^2+6r-17=0

a = 1; b = 6; c = -17;
Δ = b2-4ac
Δ = 62-4·1·(-17)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{26}}{2*1}=\frac{-6-2\sqrt{26}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{26}}{2*1}=\frac{-6+2\sqrt{26}}{2}
$