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x^2+19x=127
We move all terms to the left:
x^2+19x-(127)=0
a = 1; b = 19; c = -127;
Δ = b2-4ac
Δ = 192-4·1·(-127)
Δ = 869
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{-(19)-\sqrt{869}}{2*1}=\frac{-19-\sqrt{869}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{869}}{2*1}=\frac{-19+\sqrt{869}}{2} $
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