How do you solve #4p^2=-7p-3# using the quadratic formula?

Question

Julia Adams · Accepted Answer

See Explanation

To solve, you must first move all the terms to one side, so: #4p^2+7p+3=0#. Then, the quadratic formula is #x=\frac {-b\pm \sqrt {b^{2}-4ac}}(2a)#. Plug in 4 for a, 7 for b and 3 for c. You get #x=\frac {-7\pm \sqrt {7^{2}-4*4*3}}(2*4)=\frac {-7\pm 1}(8)# so, #x=(-7+1)/(8)=-6/8=-3/4# and #x=(-7-1)/(8)=(-8)/8=-1#. x=-1 and -3/4

Julia Adams · Answer

undefined To solve the equation $4p^2 = -7p - 3$ using the quadratic formula:

1. Rearrange the equation into the standard form $ax^2 + bx + c = 0$, where $a = 4$, $b = 7$, and $c = -3$.
2. Apply the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
3. Substitute the values of $a$, $b$, and $c$ into the quadratic formula.
4. Calculate the discriminant, $D = b^2 - 4ac$.
5. Substitute the values of $a$, $b$, and $D$ into the quadratic formula.
6. Simplify the expression.
7. Solve for $p$.
8. Determine the solutions for $p$ by considering both the positive and negative square roots.

Applying these steps:

1. $D = (-7)^2 - 4(4)(-3) = 49 + 48 = 97$.
2. $p = \frac{{-(-7) \pm \sqrt{{97}}}}{{2(4)}}$.
3. $p = \frac{{7 \pm \sqrt{{97}}}}{{8}}$.

Therefore, the solutions for $p$ are $p = \frac{{7 + \sqrt{97}}}{{8}}$ and $p = \frac{{7 - \sqrt{97}}}{{8}}$.