Question
$$3(3x+1)(2x+5)+3(2x^2+5x) = (2x+7)^2+4(6x-1)+1362$$
Answer
$$ \begin{matrix}x_1 = 8 & x_2 = -\dfrac{ 87 }{ 10 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3(3x+1)(2x+5)+3(2x^2+5x) &= (2x+7)^2+4(6x-1)+1362&& \text{simplify left and right hand side} \\[1 em](9x+3)(2x+5)+6x^2+15x &= 4x^2+28x+49+4(6x-1)+1362&& \\[1 em]18x^2+45x+6x+15+6x^2+15x &= 4x^2+28x+49+24x-4+1362&& \\[1 em]18x^2+51x+15+6x^2+15x &= 4x^2+52x+45+1362&& \\[1 em]24x^2+66x+15 &= 4x^2+52x+1407&& \text{move all terms to the left hand side } \\[1 em]24x^2+66x+15-4x^2-52x-1407 &= 0&& \text{simplify left side} \\[1 em]20x^2+14x-1392 &= 0&& \\[1 em] \end{aligned} $$
$ 20x^{2}+14x-1392 = 0 $ is a quadratic equation.
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