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4c^2-25c+21=0
a = 4; b = -25; c = +21;
Δ = b2-4ac
Δ = -252-4·4·21
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-17}{2*4}=\frac{8}{8} =1 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+17}{2*4}=\frac{42}{8} =5+1/4 $
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