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5x^2+3x=18
We move all terms to the left:
5x^2+3x-(18)=0
a = 5; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·5·(-18)
Δ = 369
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{41}}{2*5}=\frac{-3-3\sqrt{41}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{41}}{2*5}=\frac{-3+3\sqrt{41}}{10} $
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