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4x^2+19x=7x+7
We move all terms to the left:
4x^2+19x-(7x+7)=0
We get rid of parentheses
4x^2+19x-7x-7=0
We add all the numbers together, and all the variables
4x^2+12x-7=0
a = 4; b = 12; c = -7;
Δ = b2-4ac
Δ = 122-4·4·(-7)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*4}=\frac{-28}{8} =-3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*4}=\frac{4}{8} =1/2 $
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