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2x^2-15=17
We move all terms to the left:
2x^2-15-(17)=0
We add all the numbers together, and all the variables
2x^2-32=0
a = 2; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·2·(-32)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4} =4 $
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