2×(х-5)-12=3 (1-х) х-10=5 (2х+1)+57 3×(х-5)+10×(х-8)=30 2×(х+1)+3×(х-2)=24

Решите задачу:

$\\ 2x(x-5)-12=3(1-x) \\ 2x^2-10x-12=3-3x \\ 2x^2-10x-12-3+3x=0 \\ 2x^2-7x-9=0 \\ D=(-7)^2-4*2*(-9)=49+72=121 \\ \sqrt{121}=11 \\ x1= \frac{7+11}{2*2 }= \frac{20}{4}= 5 \\ x2= \frac{7-11}{4}=- \frac{4}{4}=-1$

$\\ x-10=5(2x+1)+57 \\ x-10=10x+5+57 \\ x-10x=62+10 \\ -9x=72 \\ x= \frac{72}{-9} \\ x=-8$