10x^2+24x-43=0

Solution for 10x^2+24x-43=0 equation:

10x^2+24x-43=0

a = 10; b = 24; c = -43;
Δ = b2-4ac
Δ = 242-4·10·(-43)
Δ = 2296
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:

$\sqrt{\Delta}=\sqrt{2296}=\sqrt{4*574}=\sqrt{4}*\sqrt{574}=2\sqrt{574}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{574}}{2*10}=\frac{-24-2\sqrt{574}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{574}}{2*10}=\frac{-24+2\sqrt{574}}{20}
$