-2x^2+36x=0

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Solution for -2x^2+36x=0 equation: 



-2x^2+36x=0

a = -2; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·(-2)·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*-2}=\frac{-72}{-4}
=+18
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*-2}=\frac{0}{-4}
=0
$

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