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5x^2-18+4=12x+39
We move all terms to the left:
5x^2-18+4-(12x+39)=0
We add all the numbers together, and all the variables
5x^2-(12x+39)-14=0
We get rid of parentheses
5x^2-12x-39-14=0
We add all the numbers together, and all the variables
5x^2-12x-53=0
a = 5; b = -12; c = -53;
Δ = b2-4ac
Δ = -122-4·5·(-53)
Δ = 1204
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1204}=\sqrt{4*301}=\sqrt{4}*\sqrt{301}=2\sqrt{301}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{301}}{2*5}=\frac{12-2\sqrt{301}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{301}}{2*5}=\frac{12+2\sqrt{301}}{10} $
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