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6x^2+3x^2=40
We move all terms to the left:
6x^2+3x^2-(40)=0
We add all the numbers together, and all the variables
9x^2-40=0
a = 9; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·9·(-40)
Δ = 1440
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{10}}{2*9}=\frac{0-12\sqrt{10}}{18} =-\frac{12\sqrt{10}}{18} =-\frac{2\sqrt{10}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{10}}{2*9}=\frac{0+12\sqrt{10}}{18} =\frac{12\sqrt{10}}{18} =\frac{2\sqrt{10}}{3} $
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