Question

Please help fassssst

Answer 1

Answer:

$\dfrac{9}{2}$

Step-by-step explanation:

You can do this using the quadratic equation:

$x =\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

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:

$x =\dfrac{-12\pm\sqrt{12^{2}-4(4)(-135)}}{2(4)}$

We solve for both to determine the positive one:

$x =\dfrac{-12+ \sqrt{12^{2}-4(4)(-135)}}{2(4)}$              $x =\dfrac{-12- \sqrt{12^{2}-4(4)(-135)}}{2(4)}$

$x =\dfrac{-12+ \sqrt{144+2160}}{8}$                        $x =\dfrac{-12- \sqrt{144+2160}}{8}$

$x =\dfrac{-12+ \sqrt{2304}}{8}$                                 $x =\dfrac{-12- \sqrt{2304}}{8}$

$x =\dfrac{-12+ 48}{8}$                                        $x =\dfrac{-12- 48}{8}$

$x =\dfrac{36}{8} = \dfrac{9}{2}$                                           $x =\dfrac{-60}{8} = \dfrac{-15}{2}$

So if you are looking for a positive solution, just take the positive one as x.