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3x^2+20x-32=0
a = 3; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·3·(-32)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*3}=\frac{-48}{6} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*3}=\frac{8}{6} =1+1/3 $
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