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4x^2+23x=-28
We move all terms to the left:
4x^2+23x-(-28)=0
We add all the numbers together, and all the variables
4x^2+23x+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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-9}{2*4}=\frac{-32}{8} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+9}{2*4}=\frac{-14}{8} =-1+3/4 $
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