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6x^2+3x^2=40^2
We move all terms to the left:
6x^2+3x^2-(40^2)=0
We add all the numbers together, and all the variables
9x^2-1600=0
a = 9; b = 0; c = -1600;
Δ = b2-4ac
Δ = 02-4·9·(-1600)
Δ = 57600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{57600}=240$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-240}{2*9}=\frac{-240}{18} =-13+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+240}{2*9}=\frac{240}{18} =13+1/3 $
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