Question

Need help on (x - 4)(x + 5) = 7?

Answer 1

solve both terms for x. So,
x-4=7

x+5=7

solve both for x.

Answer 2

Expand the brackets. How? Distribute EACH term.
(x - 4)(x + 5) = 7

x * x = x^2
x * 5 = 5x

-4 * x = -4x
-4 * 5 = -20

So now you have:
x^2 + 5x - 4x - 20 = 7

Combine like terms. Solve for x.

x^2 + 5x - 4x - 20 = 7

x^2 + x - 20 = 7

Subtract 7.

x^2 + 5x - 13 = 0

^That is a quadratic equation (ax^2 + bx + c = 0). Use the Quadratic formula to solve for x.

\[x = -b \pm \frac{ \sqrt{b^2 - 4ac} }{ 2a }\] \[x = -5 \pm \frac{ \sqrt{5^2 - 4(1)(-13)} }{ 2(1) }\]\[x = -5 \pm \frac{ \sqrt{25 + 52} }{ 2 }\]\[x = -5 \pm \frac{ \sqrt{77} }{ 2 } \]\[x = -5 \pm \frac{ 8.77}{ 2 }\]\[x = -5 \pm 4.385\] \[x = -5 + 4.835 =-0.615 \] \[x = -5 - 4.835 =-9.835 \]

So:
x = -0.615
and
x = -9.835

Answer 3

x^2-4x+5x-20=7
x^2+x-13=0

now find x.

Answer 4

I made a mistake when subtracting 7. Okie so...

x^2 + x - 20 = 7

Subtract 7.

x^2 + 5x - 27 = 0

^That is a quadratic equation (ax^2 + bx + c = 0). Use the Quadratic formula to solve for x.
--------------------------------------
\[x = -b \pm \frac{ \sqrt{b^2 - 4ac} }{ 2a }\]\[x = -5 \pm \frac{\sqrt{ 5^2 - 4(1)(-27)} }{ 2(1) } \]\[x = -5 \pm \frac{\sqrt{ 25 + 108)} }{ 2 }\]\[x = -5 \pm \frac{ \sqrt{133} }{ 2 }\]\[x = -5 \pm \frac{ 11.53 }{ 2 }\]\[x = -5 \pm 5.765\] \[x = -5 + 5.765 = 0.765\] \[x = -5 - 5.765 = -10.765\]

So:
x = 0.765
and
x = -10.765

Answer 5

Compare your quadratic equation with \(ax^2+bx+c=0\)
find a,b,c
then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)