2u^2+24u+70=0

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Solution for 2u^2+24u+70=0 equation:



2u^2+24u+70=0
a = 2; b = 24; c = +70;
Δ = b2-4ac
Δ = 242-4·2·70
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4}{2*2}=\frac{-28}{4} =-7 $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4}{2*2}=\frac{-20}{4} =-5 $

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