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3x^2+x^2=150
We move all terms to the left:
3x^2+x^2-(150)=0
We add all the numbers together, and all the variables
4x^2-150=0
a = 4; b = 0; c = -150;
Δ = b2-4ac
Δ = 02-4·4·(-150)
Δ = 2400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*4}=\frac{0-20\sqrt{6}}{8} =-\frac{20\sqrt{6}}{8} =-\frac{5\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*4}=\frac{0+20\sqrt{6}}{8} =\frac{20\sqrt{6}}{8} =\frac{5\sqrt{6}}{2} $
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