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5x+4x^2=18
We move all terms to the left:
5x+4x^2-(18)=0
a = 4; b = 5; c = -18;
Δ = b2-4ac
Δ = 52-4·4·(-18)
Δ = 313
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{313}}{2*4}=\frac{-5-\sqrt{313}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{313}}{2*4}=\frac{-5+\sqrt{313}}{8} $
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