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2.5x^2-10=0
a = 2.5; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·2.5·(-10)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*2.5}=\frac{-10}{5} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*2.5}=\frac{10}{5} =2 $
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