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3x^2+2x^2=36
We move all terms to the left:
3x^2+2x^2-(36)=0
We add all the numbers together, and all the variables
5x^2-36=0
a = 5; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·5·(-36)
Δ = 720
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{5}}{2*5}=\frac{0-12\sqrt{5}}{10} =-\frac{12\sqrt{5}}{10} =-\frac{6\sqrt{5}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{5}}{2*5}=\frac{0+12\sqrt{5}}{10} =\frac{12\sqrt{5}}{10} =\frac{6\sqrt{5}}{5} $
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