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3x^2+25x-50=0
a = 3; b = 25; c = -50;
Δ = b2-4ac
Δ = 252-4·3·(-50)
Δ = 1225
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-35}{2*3}=\frac{-60}{6} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+35}{2*3}=\frac{10}{6} =1+2/3 $
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