16x^2+18x-48=0

Solution for 16x^2+18x-48=0 equation:

16x^2+18x-48=0

a = 16; b = 18; c = -48;
Δ = b2-4ac
Δ = 182-4·16·(-48)
Δ = 3396
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:

$\sqrt{\Delta}=\sqrt{3396}=\sqrt{4*849}=\sqrt{4}*\sqrt{849}=2\sqrt{849}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{849}}{2*16}=\frac{-18-2\sqrt{849}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{849}}{2*16}=\frac{-18+2\sqrt{849}}{32}
$