## anonymous 5 years ago 4x^2 +24x+35=0

1. anonymous

you need to factor...

2. anonymous

x = $-24 \pm \sqrt{24^2-4(4)(35) }$ $/2(4)$

3. anonymous

good call, use the quadratic equation for that. You could, however, factor that. 35 has the following primes: 5, 7. 24 has the following primes: 2,2,2,3, and 4 has two primes, 2,2. Don't think this will factor into easy integers, so just use quadratic equation here.

4. anonymous

Doesn't look like it factors easily, so use completing the square: $4x ^2+24x +35 = 0$ (Divide by the x-squared coeficient) $x ^2+6x +35/4 = 0$ (Move the constant term to the other side) $x ^2+6x = -35/4$ (Take half the coeficient of the x term, square it, and add to both sides) $x ^2+6x +3^2 = -35/4 +3^2$ (Now the left side of the equation is a perfect square, and factors easily) $(x +3)^2 = -35/4 +9$ (take the square root of each side) $x +3 = \pm\sqrt{-35/4 +9}$ (now just move the three over, and you're done) $x =-3 \pm\sqrt{-35/4 +9}$

5. anonymous

Ends up reducing to -5/2 and -7/2