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4x^2+2x=22
We move all terms to the left:
4x^2+2x-(22)=0
a = 4; b = 2; c = -22;
Δ = b2-4ac
Δ = 22-4·4·(-22)
Δ = 356
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{89}}{2*4}=\frac{-2-2\sqrt{89}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{89}}{2*4}=\frac{-2+2\sqrt{89}}{8} $
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