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4k^2-25=0
a = 4; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·4·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*4}=\frac{-20}{8} =-2+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*4}=\frac{20}{8} =2+1/2 $
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