$\displaystyle\it\\\text{We have the equation: }\\\\ax^2+bx+c=0\\\\\text{We find the roots:}\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x_{1}=\frac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_{2}=\frac{-b-\sqrt{b^2-4ac}}{2a}\\\\\text{If } b^2-4ac>0 \text{ then }x_1, x_2\in~R\\\\\text{If } b^2-4ac=0 \text{ then }x_1=x_2\in~R\\\\\text{If } b^2-4ac$

3 0

3 years ago

Help please

exis [7]

Answer:

You can put this solution on YOUR website! 5X - 20 = 10 * X 5X - 20 = 10X Subtract 5x from each side-20 = 5X Divide each side by 5-4 = X. Let's make sure it works... 20 less than five times a number (-20-20) is equal to the product of ten and that number (10 * -4)

Step-by-step explanation:

8 0

3 years ago

What is the distance between points A and B? (Use only digits 0-9 to write the distance.)

qwelly [4]

Answer:

Step-by-step explanation:

A is on the negative side so you count the distanse as -6 -5 -4 so on and then you get to 0 and then its 12345678

3 0

3 years ago