Question

1) (2x+10) (x-1)+5(x-2)=2(7+x) 2) 1+3(2x-4)+(2x-1)(3-2x)=8

Answer 1

$(2x+10) (x-1)+5(x-2)=2(7+x)$
$2 x^2-2 x+10 x-10+5 x-10-14-2 x=0$
$2 x^2+11 x-34 = 0$
$D = b^2-4ac = 393$
$x_{1,2}= \frac{-b\pm\sqrt{D}}{2a} = \frac{-11\pm\sqrt{393}}{4} = \frac{-11\pm\sqrt{393}}{4}$

ответе:
$x_1 = \frac{-11+\sqrt{393}}{4},\; x_2 = \frac{-11-\sqrt{393}}{4}$

2) $1+3(2x-4)+(2x-1)(3-2x)=8$
$-7+6 x-12+6 x-4 x^2-3+2 x=0$
$-4 x^2+14 x-22 =0$
$D = b^2-4ac = -156$

ответ: корней нет, т.к. D<0