Question

Solve 2x^2 = 14x + 20 by using the quadratic formula?

Answer 1

\[\Large 2x^2 = 14x + 20\]


\[\Large 2x^2 - 14x - 20 = 0\]


The equation is in \(\Large ax^2 + bx + c = 0 \) form where a = 2, b = -14 and c = -20


Plug all this into the quadratic formula to get

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(-14)\pm\sqrt{(-14)^2-4(2)(-20)}}{2(2)}\]

I'll let you finish up

Answer 2

i've got up to that part. its this part i'm stuck on \[-14 \pm 2√89 divide everything by 4\]

Answer 3

The first 14 should be positive (since -(-14) = +14 = 14), keep going to get


\[\Large x = \frac{14+2\sqrt{89}}{4} \ \text{or} \ x = \frac{14-2\sqrt{89}}{4}\]


\[\Large x = \frac{2(7+\sqrt{89})}{2*2} \ \text{or} \ x = \frac{2(7-\sqrt{89})}{2*2}\]


\[\Large x = \frac{\cancel{2}(7+\sqrt{89})}{\cancel{2}*2} \ \text{or} \ x = \frac{\cancel{2}(7-\sqrt{89})}{\cancel{2}*2}\]


\[\Large x = \frac{7+\sqrt{89}}{2} \ \text{or} \ x = \frac{7-\sqrt{89}}{2}\]

Answer 4

And those are your exact answers. To get the approximate answers, use your calculator.

Answer 5

ohhhhh i get it now! thanks!

Answer 6

ok great