Question
$$3(3x+1)(2x+5)+3(2x\cdot2+5x) = (2x+7)\cdot2+4(6x-1)+1362$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 25 }{ 18 }-\dfrac{\sqrt{ 25051 }}{ 18 } & x_2 = -\dfrac{ 25 }{ 18 }+\dfrac{\sqrt{ 25051 }}{ 18 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3(3x+1)(2x+5)+3(2x\cdot2+5x) &= (2x+7)\cdot2+4(6x-1)+1362&& \text{simplify left and right hand side} \\[1 em]3(3x+1)(2x+5)+3(4x+5x) &= 4x+14+24x-4+1362&& \\[1 em]3(3x+1)(2x+5)+3\cdot9x &= 28x+10+1362&& \\[1 em](9x+3)(2x+5)+27x &= 28x+1372&& \\[1 em]18x^2+45x+6x+15+27x &= 28x+1372&& \\[1 em]18x^2+78x+15 &= 28x+1372&& \text{move all terms to the left hand side } \\[1 em]18x^2+78x+15-28x-1372 &= 0&& \text{simplify left side} \\[1 em]18x^2+50x-1357 &= 0&& \\[1 em] \end{aligned} $$
$ 18x^{2}+50x-1357 = 0 $ is a quadratic equation.
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