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4x^2+4x-15=20-19x
We move all terms to the left:
4x^2+4x-15-(20-19x)=0
We add all the numbers together, and all the variables
4x^2+4x-(-19x+20)-15=0
We get rid of parentheses
4x^2+4x+19x-20-15=0
We add all the numbers together, and all the variables
4x^2+23x-35=0
a = 4; b = 23; c = -35;
Δ = b2-4ac
Δ = 232-4·4·(-35)
Δ = 1089
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{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*4}=\frac{-56}{8} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*4}=\frac{10}{8} =1+1/4 $
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