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-3a^2+39a-90=0
a = -3; b = 39; c = -90;
Δ = b2-4ac
Δ = 392-4·(-3)·(-90)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-21}{2*-3}=\frac{-60}{-6} =+10 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+21}{2*-3}=\frac{-18}{-6} =+3 $
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