ПОЖ... помогите x(x+3)=10 x(x-1)=42 (x+7)x=18 (x-9)x=36 ТОлько пишите ясно прошу

X(x+3)=10
$x^{2}$ +3x=10
$x^{2} +3x-10=0 \\ D= b^{2} -4ac=9-4*(-10)=9+40=49 \\ x_{1} = \frac{ -b + \sqrt{D} }{2a} = \frac{-3+7}{2} =2 \\x_{2} = \frac{ -b -\sqrt{D} }{2a} = \frac{-3-7}{2} =-5$
x(x-1)=42
$x^{2} -x-42=0 \\ D=1+42*4=1+168=169 \\ x_{1} = \frac{ -b + \sqrt{D} }{2a} = \frac{-1+13}{2}=6 \\x_{2} = \frac{ -b -\sqrt{D} }{2a} = \frac{-1-13}{2} = \frac{-14}{2} =-7$
(x+7)x=18
$x^{2} +7x-18=0 \\ D=49+4*18=49+72=121 \\ x_{1} = \frac{ -b + \sqrt{D} }{2a} = \frac{-7+11}{2}=2 \\ x_{2} = \frac{- b - \sqrt{D} }{2a} = \frac{-7-11}{2}=-9$
(x-9)x=36
$x^{2} -9x-36=0 \\ D=81+36*4=81+144=225 \\ x_{1} = \frac{ -b + \sqrt{D} }{2a} = \frac{9+15}{2}=12 \\ x_{2} = \frac{ -b - \sqrt{D} }{2a} = \frac{9-15}{2}=-3$