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x^2+14x=48
We move all terms to the left:
x^2+14x-(48)=0
a = 1; b = 14; c = -48;
Δ = b2-4ac
Δ = 142-4·1·(-48)
Δ = 388
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{97}}{2*1}=\frac{-14-2\sqrt{97}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{97}}{2*1}=\frac{-14+2\sqrt{97}}{2} $
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