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(x+4)(2x-7)=63^2
We move all terms to the left:
(x+4)(2x-7)-(63^2)=0
We add all the numbers together, and all the variables
(x+4)(2x-7)-3969=0
We multiply parentheses ..
(+2x^2-7x+8x-28)-3969=0
We get rid of parentheses
2x^2-7x+8x-28-3969=0
We add all the numbers together, and all the variables
2x^2+x-3997=0
a = 2; b = 1; c = -3997;
Δ = b2-4ac
Δ = 12-4·2·(-3997)
Δ = 31977
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{31977}=\sqrt{9*3553}=\sqrt{9}*\sqrt{3553}=3\sqrt{3553}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{3553}}{2*2}=\frac{-1-3\sqrt{3553}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{3553}}{2*2}=\frac{-1+3\sqrt{3553}}{4} $
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