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45x^2+18x-10=0
a = 45; b = 18; c = -10;
Δ = b2-4ac
Δ = 182-4·45·(-10)
Δ = 2124
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2124}=\sqrt{36*59}=\sqrt{36}*\sqrt{59}=6\sqrt{59}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{59}}{2*45}=\frac{-18-6\sqrt{59}}{90} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{59}}{2*45}=\frac{-18+6\sqrt{59}}{90} $
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