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12x^2+20x-400=0
a = 12; b = 20; c = -400;
Δ = b2-4ac
Δ = 202-4·12·(-400)
Δ = 19600
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{19600}=140$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-140}{2*12}=\frac{-160}{24} =-6+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+140}{2*12}=\frac{120}{24} =5 $
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