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3x^2+12x+13=4x^2+25
We move all terms to the left:
3x^2+12x+13-(4x^2+25)=0
We get rid of parentheses
3x^2-4x^2+12x-25+13=0
We add all the numbers together, and all the variables
-1x^2+12x-12=0
a = -1; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·(-1)·(-12)
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{6}}{2*-1}=\frac{-12-4\sqrt{6}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{6}}{2*-1}=\frac{-12+4\sqrt{6}}{-2} $
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