y^2+y-92=0

Solution for y^2+y-92=0 equation:

y^2+y-92=0

a = 1; b = 1; c = -92;
Δ = b2-4ac
Δ = 12-4·1·(-92)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{41}}{2*1}=\frac{-1-3\sqrt{41}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{41}}{2*1}=\frac{-1+3\sqrt{41}}{2}
$