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-2a^2+10a+48=0
a = -2; b = 10; c = +48;
Δ = b2-4ac
Δ = 102-4·(-2)·48
Δ = 484
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{484}=22$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*-2}=\frac{-32}{-4} =+8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*-2}=\frac{12}{-4} =-3 $
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