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x^2-3x-8=10-3x
We move all terms to the left:
x^2-3x-8-(10-3x)=0
We add all the numbers together, and all the variables
x^2-3x-(-3x+10)-8=0
We get rid of parentheses
x^2-3x+3x-10-8=0
We add all the numbers together, and all the variables
x^2-18=0
a = 1; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·1·(-18)
Δ = 72
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*1}=\frac{0-6\sqrt{2}}{2} =-\frac{6\sqrt{2}}{2} =-3\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*1}=\frac{0+6\sqrt{2}}{2} =\frac{6\sqrt{2}}{2} =3\sqrt{2} $
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