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25x^2-35x-12=0
a = 25; b = -35; c = -12;
Δ = b2-4ac
Δ = -352-4·25·(-12)
Δ = 2425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2425}=\sqrt{25*97}=\sqrt{25}*\sqrt{97}=5\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{97}}{2*25}=\frac{35-5\sqrt{97}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{97}}{2*25}=\frac{35+5\sqrt{97}}{50} $
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