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4a^2-23a+28=0
a = 4; b = -23; c = +28;
Δ = b2-4ac
Δ = -232-4·4·28
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-9}{2*4}=\frac{14}{8} =1+3/4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+9}{2*4}=\frac{32}{8} =4 $
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