Question

use the quadratic formula to solve the equation
5x^2-4x-3=0

Answer 1

x(5x-4)=3

Answer 2

how did you get that?

Answer 3

quadratic equation

Answer 4

can you explain it please?

Answer 5

you have to know the quadratic formula before you can use it ....

Answer 6

@DoomDude use quadratic formula not factoring

Answer 7

the quadratic formula is the aftereffects of completing the square

Answer 8

can someone explain to me how to use the quadratic formula by using this problem?

Answer 9

step by step

Answer 10

\[ax^2+bx+c=0\]
\[ax^2+bx=-c\]
\[x^2+\frac{b}{a}x=-\frac{c}{a}\]
\[x^2+\frac{b}{a}x+(\frac {b}{2a})^2=-\frac ca+(\frac{b}{2a})^2\]
\[(x+\frac{b}{2a})^2=-\frac ca+\frac{b^2}{4a^2}\]
\[(x+\frac{b}{2a})^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\]
\[x+\frac{b}{2a}=\pm\sqrt{-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}}\]
\[x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\]
\[x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 11

bravo, very thorough

Answer 12

and if you dont want to go thru all the effort of completeing a square; just jump to the end of it and plug in