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4y^2-29y+25=0
a = 4; b = -29; c = +25;
Δ = b2-4ac
Δ = -292-4·4·25
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-21}{2*4}=\frac{8}{8} =1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+21}{2*4}=\frac{50}{8} =6+1/4 $
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