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25x^2-4=0
a = 25; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·25·(-4)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*25}=\frac{-20}{50} =-2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*25}=\frac{20}{50} =2/5 $
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