Answer:
[tex]\dfrac{9}{2}[/tex]
Step-by-step explanation:
You can do this using the quadratic equation:
[tex]x =\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
Let's first setup our expression:
4x² + 12x = 135
the quadratic formula is:
ax² + bx + c = 0
4x² + 12x - 135 = 0
Then:
a = 4
b = 12
c = -135
Now we plug in our coefficients and solve:
[tex]x =\dfrac{-12\pm\sqrt{12^{2}-4(4)(-135)}}{2(4)}[/tex]
We solve for both to determine the positive one:
[tex]x =\dfrac{-12+ \sqrt{12^{2}-4(4)(-135)}}{2(4)}[/tex] [tex]x =\dfrac{-12- \sqrt{12^{2}-4(4)(-135)}}{2(4)}[/tex]
[tex]x =\dfrac{-12+ \sqrt{144+2160}}{8}[/tex] [tex]x =\dfrac{-12- \sqrt{144+2160}}{8}[/tex]
[tex]x =\dfrac{-12+ \sqrt{2304}}{8}[/tex] [tex]x =\dfrac{-12- \sqrt{2304}}{8}[/tex]
[tex]x =\dfrac{-12+ 48}{8}[/tex] [tex]x =\dfrac{-12- 48}{8}[/tex]
[tex]x =\dfrac{36}{8} = \dfrac{9}{2}[/tex] [tex]x =\dfrac{-60}{8} = \dfrac{-15}{2}[/tex]
So if you are looking for a positive solution, just take the positive one as x.
