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6x^2+35x-25=0
a = 6; b = 35; c = -25;
Δ = b2-4ac
Δ = 352-4·6·(-25)
Δ = 1825
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{73}}{2*6}=\frac{-35-5\sqrt{73}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{73}}{2*6}=\frac{-35+5\sqrt{73}}{12} $
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