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3b^2+10b-125=0
a = 3; b = 10; c = -125;
Δ = b2-4ac
Δ = 102-4·3·(-125)
Δ = 1600
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{1600}=40$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-40}{2*3}=\frac{-50}{6} =-8+1/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+40}{2*3}=\frac{30}{6} =5 $
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