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(X-3)(X+4)=18+X
We move all terms to the left:
(X-3)(X+4)-(18+X)=0
We add all the numbers together, and all the variables
(X-3)(X+4)-(X+18)=0
We get rid of parentheses
(X-3)(X+4)-X-18=0
We multiply parentheses ..
(+X^2+4X-3X-12)-X-18=0
We add all the numbers together, and all the variables
(+X^2+4X-3X-12)-1X-18=0
We get rid of parentheses
X^2+4X-3X-1X-12-18=0
We add all the numbers together, and all the variables
X^2-30=0
a = 1; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·1·(-30)
Δ = 120
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*1}=\frac{0-2\sqrt{30}}{2} =-\frac{2\sqrt{30}}{2} =-\sqrt{30} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*1}=\frac{0+2\sqrt{30}}{2} =\frac{2\sqrt{30}}{2} =\sqrt{30} $
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