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(3x+4)(7x+7)=161
We move all terms to the left:
(3x+4)(7x+7)-(161)=0
We multiply parentheses ..
(+21x^2+21x+28x+28)-161=0
We get rid of parentheses
21x^2+21x+28x+28-161=0
We add all the numbers together, and all the variables
21x^2+49x-133=0
a = 21; b = 49; c = -133;
Δ = b2-4ac
Δ = 492-4·21·(-133)
Δ = 13573
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{13573}=\sqrt{49*277}=\sqrt{49}*\sqrt{277}=7\sqrt{277}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-7\sqrt{277}}{2*21}=\frac{-49-7\sqrt{277}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+7\sqrt{277}}{2*21}=\frac{-49+7\sqrt{277}}{42} $
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