$10x^2=29x-21 \\
10x^2-29x+21=0\\
\Delta=(-29)^2-4\cdot10\cdot21=841-840=1\\
\sqrt{\Delta}=\sqrt1=1\\
x_1=\frac{-(-29)-1}{2\cdot10}=\frac{28}{20}=\frac{7}{5}\\
x_2=\frac{-(-29)+1}{2\cdot10}=\frac{30}{20}=\frac{3}{2}\\$

4 0

3 years ago

Solve. x^2 8x 25 = 0

kykrilka [37]

This equation is unsolvable.
x^2+8x+16=0
(x+4)^2+9=0
(x+4)^2= -9
anything to the power of whatever has to be greater than -1, it means it can be 0 or more but never a negative number. If it is a negative number, then it is false.

6 0

4 years ago

X + 2y = 7<br>x - y = 1<br><br>Solve for x and y?

34kurt

Answer:

y=2

x=3

Step-by-step explanation:

make x - y = 1 into x= y + 1

put that wherever x is in the other equation

so y + 1 + 2y =7. then combine like terms into 3y =6. then divide 6 by 3 so you get y = 2

then substitute 2 into the y in the equation x - y = 1 and you get x = 3

7 0

3 years ago

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