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6a^2+35a+25=0
a = 6; b = 35; c = +25;
Δ = b2-4ac
Δ = 352-4·6·25
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-25}{2*6}=\frac{-60}{12} =-5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+25}{2*6}=\frac{-10}{12} =-5/6 $
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