3x2+16x-48=0

Solution for 3x2+16x-48=0 equation:

3x^2+16x-48=0

a = 3; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·3·(-48)
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{13}}{2*3}=\frac{-16-8\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{13}}{2*3}=\frac{-16+8\sqrt{13}}{6}
$