Решить уравнение1)х^2-7х=02)4х^2-169=03)6х^2+х-2=04)3х^2+6х=8х^2-15х

1)х²-7х=0
x(x-7)=0
x=0 x-7=0
x=7
2)4х²-169=0
4x²=169
x²= $\frac{169}{4}$ >0
$x_{1} = \sqrt{ \frac{169}{4} } = \frac{13}{2} =6 \frac{1}{2} =6,5$
$x_{2} =- \sqrt{ \frac{169}{4} } = -\frac{13}{2} =-6 \frac{1}{2} =-6,5$
3)6х²+х-2=0
D=1²-4*6*(-2)=49=7²>0
$x_{1} = \frac{1+ \sqrt{7 ^{2} } }{2*6} = \frac{8}{12} = \frac{2}{3}$
$x_{2} = \frac{1- \sqrt{7 ^{2} } }{2*6} = \frac{-6}{12} = -\frac{1}{2} =-0,5$

4)3х²+6х=8х²-15х
-5x²+21x=0
x(-5x+21)=0
x=0 -5x+21=0
-5x=-21
x=4,2