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-3u^2+20u+63=0
a = -3; b = 20; c = +63;
Δ = b2-4ac
Δ = 202-4·(-3)·63
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1156}=34$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-34}{2*-3}=\frac{-54}{-6} =+9 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+34}{2*-3}=\frac{14}{-6} =-2+1/3 $
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