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4x^2-19x+23=4x+8
We move all terms to the left:
4x^2-19x+23-(4x+8)=0
We get rid of parentheses
4x^2-19x-4x-8+23=0
We add all the numbers together, and all the variables
4x^2-23x+15=0
a = 4; b = -23; c = +15;
Δ = b2-4ac
Δ = -232-4·4·15
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-17}{2*4}=\frac{6}{8} =3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+17}{2*4}=\frac{40}{8} =5 $
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