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3x^2-50=2x^2+75
We move all terms to the left:
3x^2-50-(2x^2+75)=0
We get rid of parentheses
3x^2-2x^2-75-50=0
We add all the numbers together, and all the variables
x^2-125=0
a = 1; b = 0; c = -125;
Δ = b2-4ac
Δ = 02-4·1·(-125)
Δ = 500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{5}}{2*1}=\frac{0-10\sqrt{5}}{2} =-\frac{10\sqrt{5}}{2} =-5\sqrt{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{5}}{2*1}=\frac{0+10\sqrt{5}}{2} =\frac{10\sqrt{5}}{2} =5\sqrt{5} $
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