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2x^2+x^2=25^2
We move all terms to the left:
2x^2+x^2-(25^2)=0
We add all the numbers together, and all the variables
3x^2-625=0
a = 3; b = 0; c = -625;
Δ = b2-4ac
Δ = 02-4·3·(-625)
Δ = 7500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7500}=\sqrt{2500*3}=\sqrt{2500}*\sqrt{3}=50\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50\sqrt{3}}{2*3}=\frac{0-50\sqrt{3}}{6} =-\frac{50\sqrt{3}}{6} =-\frac{25\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50\sqrt{3}}{2*3}=\frac{0+50\sqrt{3}}{6} =\frac{50\sqrt{3}}{6} =\frac{25\sqrt{3}}{3} $
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