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4=-16x^2+18x
We move all terms to the left:
4-(-16x^2+18x)=0
We get rid of parentheses
16x^2-18x+4=0
a = 16; b = -18; c = +4;
Δ = b2-4ac
Δ = -182-4·16·4
Δ = 68
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{17}}{2*16}=\frac{18-2\sqrt{17}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{17}}{2*16}=\frac{18+2\sqrt{17}}{32} $
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