7y^2+34y-133=0

Solution for 7y^2+34y-133=0 equation:

7y^2+34y-133=0

a = 7; b = 34; c = -133;
Δ = b2-4ac
Δ = 342-4·7·(-133)
Δ = 4880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4880}=\sqrt{16*305}=\sqrt{16}*\sqrt{305}=4\sqrt{305}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-4\sqrt{305}}{2*7}=\frac{-34-4\sqrt{305}}{14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+4\sqrt{305}}{2*7}=\frac{-34+4\sqrt{305}}{14}
$