$$3(x-4)^{2}+5(x-3)^{2}=(2x-5)(4x-1)-40$$
$x=4$
$$3\left(x^{2}-8x+16\right)+5\left(x-3\right)^{2}=\left(2x-5\right)\left(4x-1\right)-40$$
$$3x^{2}-24x+48+5\left(x-3\right)^{2}=\left(2x-5\right)\left(4x-1\right)-40$$
$$3x^{2}-24x+48+5\left(x^{2}-6x+9\right)=\left(2x-5\right)\left(4x-1\right)-40$$
$$3x^{2}-24x+48+5x^{2}-30x+45=\left(2x-5\right)\left(4x-1\right)-40$$
$$8x^{2}-24x+48-30x+45=\left(2x-5\right)\left(4x-1\right)-40$$
$$8x^{2}-54x+48+45=\left(2x-5\right)\left(4x-1\right)-40$$
$$8x^{2}-54x+93=\left(2x-5\right)\left(4x-1\right)-40$$
$$8x^{2}-54x+93=8x^{2}-22x+5-40$$
$$8x^{2}-54x+93=8x^{2}-22x-35$$
$$8x^{2}-54x+93-8x^{2}=-22x-35$$
$$-54x+93=-22x-35$$
$$-54x+93+22x=-35$$
$$-32x+93=-35$$
$$-32x=-35-93$$
$$-32x=-128$$
$$x=\frac{-128}{-32}$$
$$x=4$$
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