-16x^2+36=0

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Solution for -16x^2+36=0 equation:



-16x^2+36=0
a = -16; b = 0; c = +36;
Δ = b2-4ac
Δ = 02-4·(-16)·36
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*-16}=\frac{-48}{-32} =1+1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-16}=\frac{48}{-32} =-1+1/2 $

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