9x^2+10x-43=0

Solution for 9x^2+10x-43=0 equation:

9x^2+10x-43=0

a = 9; b = 10; c = -43;
Δ = b2-4ac
Δ = 102-4·9·(-43)
Δ = 1648
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:

$\sqrt{\Delta}=\sqrt{1648}=\sqrt{16*103}=\sqrt{16}*\sqrt{103}=4\sqrt{103}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{103}}{2*9}=\frac{-10-4\sqrt{103}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{103}}{2*9}=\frac{-10+4\sqrt{103}}{18}
$