(2x − 5)(x + 2) = 18
$2x^2-5x+4x-10-18=0$
$2x^2-x-28=0$
$D=b^2-4ac=(-1{)}^2-4\ast 2\ast (-28)=1+224=225>0$
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{1+\sqrt{225}}{2\ast 2}=\frac{1+15}{4}=\frac{16}{4}=4$
$x_2=\frac{-b-\sqrt{D}}{2a}=\frac{1-\sqrt{225}}{2\ast 2}=\frac{1-15}{4}=\frac{-14}{4}=\frac{-7}{2}=-3,5$
Ответ: при x = −3,5 и x = 4

$(4x-3{)}^2+(3x-1)(3x+1)=9$
$16x^2-24x+9+9x^2-1-9=0$
$25x^2-24x-1=0$
$D=b^2-4ac=(-24{)}^2-4\ast 25\ast (-1)=576+100=676>0$
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{24+\sqrt{676}}{2\ast 25}=\frac{24+26}{50}=\frac{50}{50}=1$
$x_2=\frac{-b-\sqrt{D}}{2a}=\frac{24-\sqrt{676}}{2\ast 25}=\frac{24-26}{50}=\frac{-2}{50}=-\frac{1}{25}$
Ответ: при $x=-\frac{1}{25}$ и x = 1

$(x+3{)}^2-(2x-1{)}^2=16$
$x^2+6x+9-(4x^2-4x+1)=16$
$x^2+6x+9-4x^2+4x-1-16=0$
$-3x^2+10x-8=0$
$D=b^2-4ac={10}^2-4\ast (-3)\ast (-8)=100-96=4>0$
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{-10+\sqrt{4}}{2\ast (-3)}=\frac{-10+2}{-6}=\frac{-8}{-6}=\frac{4}{3}=1\frac{1}{3}$
$x_2=\frac{-b-\sqrt{D}}{2a}=\frac{-10-\sqrt{4}}{2\ast (-3)}=\frac{-10-2}{-6}=\frac{-12}{-6}=2$
Ответ: при $x=1\frac{1}{3}$ и x = 2

$(x-6{)}^2-2x(x+3)=30-12x$
$x^2-12x+36-2x^2-6x-30+12x=0$
$-x^2-6x+6=0$
$D=b^2-4ac=(-6{)}^2-4\ast (-1)\ast 6=36+24=60>0$
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{6+\sqrt{60}}{2\ast (-1)}=\frac{6+\sqrt{4\ast 15}}{-2}=\frac{6+2\sqrt{15}}{-2}=-\frac{2(3+\sqrt{15})}{2}=-3-\sqrt{15}$
$x_2=\frac{-b-\sqrt{D}}{2a}=\frac{6-\sqrt{60}}{2\ast (-1)}=\frac{6-\sqrt{4\ast 15}}{-2}=\frac{6-2\sqrt{15}}{-2}=-\frac{2(3-\sqrt{15})}{2}=-3+\sqrt{15}$
Ответ: при $x=-3-\sqrt{15}$ и $x=-3+\sqrt{15}$

(x + 7)(x − 8) − (4x + 1)(x − 2) = −21x
$x^2+7x-8x-56-(4x^2+x-8x-2)+21x=0$
$x^2-x-56-4x^2-x+8x+2+21x=0$
$-3x^2+27x-54=0$ |: (−3)
$x^2-9x+18=0$
$D=b^2-4ac=(-9{)}^2-4\ast 1\ast 18=81-72=9>0$
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{9+\sqrt{9}}{2\ast 1}=\frac{9+3}{2}=\frac{12}{2}=6$
$x_2=\frac{-b-\sqrt{D}}{2a}=\frac{9-\sqrt{9}}{2\ast 1}=\frac{9-3}{2}=\frac{6}{2}=3$
Ответ: при x = 3 и x = 6

(2x − 1)(2x + 1) − x(1 − x) = 2x(x + 1)
$4x^2-1-x+x^2=2x^2+2x$
$5x^2-x-1-2x^2-2x=0$
$3x^2-3x-1=0$
$D=b^2-4ac=(-3{)}^2-4\ast 3\ast (-1)=9+12=21>0$
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{3+\sqrt{21}}{2\ast 3}=\frac{3+\sqrt{21}}{6}$
$x_2=\frac{-b-\sqrt{D}}{2a}=\frac{3-\sqrt{21}}{2\ast 3}=\frac{3-\sqrt{21}}{6}$
Ответ: при $x=\frac{3-\sqrt{21}}{6}$ и $x=\frac{3+\sqrt{21}}{6}$