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3x^2-91x+400=0
a = 3; b = -91; c = +400;
Δ = b2-4ac
Δ = -912-4·3·400
Δ = 3481
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{3481}=59$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-91)-59}{2*3}=\frac{32}{6} =5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-91)+59}{2*3}=\frac{150}{6} =25 $
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