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3x^2+40x-400=0
a = 3; b = 40; c = -400;
Δ = b2-4ac
Δ = 402-4·3·(-400)
Δ = 6400
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{6400}=80$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-80}{2*3}=\frac{-120}{6} =-20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+80}{2*3}=\frac{40}{6} =6+2/3 $
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