# How do you solve #x^2+28x+196=0#?

##### 3 Answers

#### Explanation:

#x^2+28x+196" is a "color(blue)"perfect square"#

#"note that "(a+b)^2=a^2+2ab+b^2#

#"with "a=x" and "b=14#

#rArrx^2+28x+196=(x+14)^2#

#"solve "(x+14)^2=0#

#rArrx+14=0rArrx=-14" (repeated)"#

Either by inspection, graphing, or the quadratic formula.

#### Explanation:

**Solve by Graphing**

Every quadratic is a function on a graph. Just replace 0 with y= and graph the quadratic as a function. the x-coordinate(s) wherever the function intersects the x axis are your solution(s). in this instance, there's only 1 x-intercept and therefore only one solution.

**Solve by Inspection**

Ask yourself, what factors of

Think, if

Then factor out the quadratic into 2 binomials...

.. and rearrange each individually to equal

**Quadratic formula**

If the quadratic doesn't look easy to factor, then pull out the quadratic formula!

Hope it helps!

The solution of this given quadratic equation is

#### Explanation:

In the given expression we see the polynomials have the following numbers as their coefficients

In middle term factorisation we have to split lower degree polynomial into two in our

Here those two factors are

=

=

=

=

I hope this was helpful. *:)*