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3x^2+50=5x^2
We move all terms to the left:
3x^2+50-(5x^2)=0
We add all the numbers together, and all the variables
-2x^2+50=0
a = -2; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-2)·50
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-2}=\frac{-20}{-4} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-2}=\frac{20}{-4} =-5 $
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