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132x^2+10x-25=0
a = 132; b = 10; c = -25;
Δ = b2-4ac
Δ = 102-4·132·(-25)
Δ = 13300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{13300}=\sqrt{100*133}=\sqrt{100}*\sqrt{133}=10\sqrt{133}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{133}}{2*132}=\frac{-10-10\sqrt{133}}{264} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{133}}{2*132}=\frac{-10+10\sqrt{133}}{264} $
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