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10x^2-10=480
We move all terms to the left:
10x^2-10-(480)=0
We add all the numbers together, and all the variables
10x^2-490=0
a = 10; b = 0; c = -490;
Δ = b2-4ac
Δ = 02-4·10·(-490)
Δ = 19600
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{19600}=140$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140}{2*10}=\frac{-140}{20} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140}{2*10}=\frac{140}{20} =7 $
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