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25+2x^2=3x^2
We move all terms to the left:
25+2x^2-(3x^2)=0
We add all the numbers together, and all the variables
-1x^2+25=0
a = -1; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·(-1)·25
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-1}=\frac{10}{-2} =-5 $
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