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4x^2=3x^2+13
We move all terms to the left:
4x^2-(3x^2+13)=0
We get rid of parentheses
4x^2-3x^2-13=0
We add all the numbers together, and all the variables
x^2-13=0
a = 1; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·1·(-13)
Δ = 52
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{13}}{2*1}=\frac{0-2\sqrt{13}}{2} =-\frac{2\sqrt{13}}{2} =-\sqrt{13} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{13}}{2*1}=\frac{0+2\sqrt{13}}{2} =\frac{2\sqrt{13}}{2} =\sqrt{13} $
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