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4x+x^2=150
We move all terms to the left:
4x+x^2-(150)=0
a = 1; b = 4; c = -150;
Δ = b2-4ac
Δ = 42-4·1·(-150)
Δ = 616
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{616}=\sqrt{4*154}=\sqrt{4}*\sqrt{154}=2\sqrt{154}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{154}}{2*1}=\frac{-4-2\sqrt{154}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{154}}{2*1}=\frac{-4+2\sqrt{154}}{2} $
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