7r2+17r-134=8

Solution for 7r2+17r-134=8 equation:

7r^2+17r-134=8

We move all terms to the left:

7r^2+17r-134-(8)=0

We add all the numbers together, and all the variables

7r^2+17r-142=0

a = 7; b = 17; c = -142;
Δ = b2-4ac
Δ = 172-4·7·(-142)
Δ = 4265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{4265}}{2*7}=\frac{-17-\sqrt{4265}}{14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{4265}}{2*7}=\frac{-17+\sqrt{4265}}{14}
$