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50=x2+12
We move all terms to the left:
50-(x2+12)=0
We add all the numbers together, and all the variables
-(+x^2+12)+50=0
We get rid of parentheses
-x^2-12+50=0
We add all the numbers together, and all the variables
-1x^2+38=0
a = -1; b = 0; c = +38;
Δ = b2-4ac
Δ = 02-4·(-1)·38
Δ = 152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{38}}{2*-1}=\frac{0-2\sqrt{38}}{-2} =-\frac{2\sqrt{38}}{-2} =-\frac{\sqrt{38}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{38}}{2*-1}=\frac{0+2\sqrt{38}}{-2} =\frac{2\sqrt{38}}{-2} =\frac{\sqrt{38}}{-1} $
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