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2x^2+4x^2=10
We move all terms to the left:
2x^2+4x^2-(10)=0
We add all the numbers together, and all the variables
6x^2-10=0
a = 6; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·6·(-10)
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*6}=\frac{0-4\sqrt{15}}{12} =-\frac{4\sqrt{15}}{12} =-\frac{\sqrt{15}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*6}=\frac{0+4\sqrt{15}}{12} =\frac{4\sqrt{15}}{12} =\frac{\sqrt{15}}{3} $
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