-3x2+8x+16=0

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Solution for -3x2+8x+16=0 equation:



-3x^2+8x+16=0
a = -3; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·(-3)·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*-3}=\frac{-24}{-6} =+4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*-3}=\frac{8}{-6} =-1+1/3 $

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