Question

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

Answer 1

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
$x^{2} -9x-36=0 \ D=81+36*4=81+144=225 \ x_{1} = frac{ -b + sqrt{D} }{2a} = frac{9+15}{2}=12 \ x_{2} = frac{ -b - sqrt{D} }{2a} = frac{9-15}{2}=-3$