y^2-10y-48=0

Solution for y^2-10y-48=0 equation:

y^2-10y-48=0

a = 1; b = -10; c = -48;
Δ = b2-4ac
Δ = -102-4·1·(-48)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{73}}{2*1}=\frac{10-2\sqrt{73}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{73}}{2*1}=\frac{10+2\sqrt{73}}{2}
$