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x2-18x-48=0
We add all the numbers together, and all the variables
x^2-18x-48=0
a = 1; b = -18; c = -48;
Δ = b2-4ac
Δ = -182-4·1·(-48)
Δ = 516
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{129}}{2*1}=\frac{18-2\sqrt{129}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{129}}{2*1}=\frac{18+2\sqrt{129}}{2} $
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