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3x^2+10=110+2x^2
We move all terms to the left:
3x^2+10-(110+2x^2)=0
We get rid of parentheses
3x^2-2x^2-110+10=0
We add all the numbers together, and all the variables
x^2-100=0
a = 1; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·1·(-100)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*1}=\frac{-20}{2} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*1}=\frac{20}{2} =10 $
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