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4x^2-2.56=0
a = 4; b = 0; c = -2.56;
Δ = b2-4ac
Δ = 02-4·4·(-2.56)
Δ = 40.96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{40.96}}{2*4}=\frac{0-\sqrt{40.96}}{8} =-\frac{\sqrt{}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{40.96}}{2*4}=\frac{0+\sqrt{40.96}}{8} =\frac{\sqrt{}}{8} $
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