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x^2+2x^2=5.3^2
We move all terms to the left:
x^2+2x^2-(5.3^2)=0
We add all the numbers together, and all the variables
3x^2-28.09=0
a = 3; b = 0; c = -28.09;
Δ = b2-4ac
Δ = 02-4·3·(-28.09)
Δ = 337.08
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{337.08}}{2*3}=\frac{0-\sqrt{337.08}}{6} =-\frac{\sqrt{}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{337.08}}{2*3}=\frac{0+\sqrt{337.08}}{6} =\frac{\sqrt{}}{6} $
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