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(2x-3)(x+7)=11x+11=
We move all terms to the left:
(2x-3)(x+7)-(11x+11)=0
We get rid of parentheses
(2x-3)(x+7)-11x-11=0
We multiply parentheses ..
(+2x^2+14x-3x-21)-11x-11=0
We get rid of parentheses
2x^2+14x-3x-11x-21-11=0
We add all the numbers together, and all the variables
2x^2-32=0
a = 2; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·2·(-32)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4} =4 $
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