How would you solve this using the quadratic formula?

Question

-x^2+3x+5x=-6+5x

THis has to =0 but taking the 5x from the right to the left side makes the x terms=0. How is this question done I am unsure how to approach this.

Answer 1

-x^2 + 3x + 5x = -6 + 5x (Remove all terms to one side of '=' sign).

x^2 - 3x - 5x -6 + 5x = 0 (Collect like terms)
x^2 - 3x - 6 = 0 (use quadratic formula)

x = --3 +/- sq rt[ (-3)^2 - 4(1)(-6)] / 2(1)
x = 3 +/- sq rt[9 + 24] / 2
x = 3 +/- sq rt[33] / 2
x = 3 +/-5.744562647 / 2
x = 3 + 5.744562647/2
x = 8.744562647/2
x = 4.372281323

&
x = 3 - 5.744562647/2
x = - 2.744562647 / 2
x = -1.372281324

Answer 2

yes the 5x terms subtract out but you still have the 3x so b = 3 in th formula, a = -1 and c = 6 (after you take the 6 over)

Note: even if the x terms go to 0, that just means you let b = 0 in the formula

Answer 3

Actually because there is an additional 3x moving the 5x from the right to the left yields 3x and not 0. But regardless, the quadratic formula is
[-b+- sq root (b^2 - 4*a*c)] / 2*a

and so your roots would become [-3+- sq root ( 3^2 - 4(-1)(6))] / 2(-1)
or -3 +- sq root(9+24) / (-2)
[-3 + sq root(33)] / (-2) and [-3 - sq root(33)] / (-2)
or -1.3723 and 4.3723

Answer 4

-x^2 + 3x + 5x = -6 +5x
-x^2 + 3x + 5x + 6 - 5x=0
-x^2 + 3x +6=0 D=9 - 4*6*(-1)= 9 + 24 = 33
so x=(-3±√33)/2*(-1)=(3±√33)/2
so x= (3 + √33)/2 or x= (3 - √33)/2

Answer 5

If you subtract, 5x and 6, the equation would be -X^2+ 3X+6=0

Then you can use the quadratic formula.

Answer 6

-x^2+3x+6=0 the equation reduces down to this.

a=-1, b=3, c=6
plug them in.

Answer 7

x^2-3x-6 = 0
x = (3±√33)/2