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9x^2+2x-43=0
a = 9; b = 2; c = -43;
Δ = b2-4ac
Δ = 22-4·9·(-43)
Δ = 1552
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{97}}{2*9}=\frac{-2-4\sqrt{97}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{97}}{2*9}=\frac{-2+4\sqrt{97}}{18} $
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