2x^2+16x-43=0

Solution for 2x^2+16x-43=0 equation:

2x^2+16x-43=0

a = 2; b = 16; c = -43;
Δ = b2-4ac
Δ = 162-4·2·(-43)
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-10\sqrt{6}}{2*2}=\frac{-16-10\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+10\sqrt{6}}{2*2}=\frac{-16+10\sqrt{6}}{4}
$