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2x^2+2x+16=7x+19
We move all terms to the left:
2x^2+2x+16-(7x+19)=0
We get rid of parentheses
2x^2+2x-7x-19+16=0
We add all the numbers together, and all the variables
2x^2-5x-3=0
a = 2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·2·(-3)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*2}=\frac{-2}{4} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*2}=\frac{12}{4} =3 $
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