0=-16x2+32x+48

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Solution for 0=-16x2+32x+48 equation:



0=-16x^2+32x+48
We move all terms to the left:
0-(-16x^2+32x+48)=0
We add all the numbers together, and all the variables
-(-16x^2+32x+48)=0
We get rid of parentheses
16x^2-32x-48=0
a = 16; b = -32; c = -48;
Δ = b2-4ac
Δ = -322-4·16·(-48)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-64}{2*16}=\frac{-32}{32} =-1 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+64}{2*16}=\frac{96}{32} =3 $

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