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12b^2+35b+25=0
a = 12; b = 35; c = +25;
Δ = b2-4ac
Δ = 352-4·12·25
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5}{2*12}=\frac{-40}{24} =-1+2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5}{2*12}=\frac{-30}{24} =-1+1/4 $
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