How do you solve #x^2+5x-6=0# using the quadratic formula?

3 Answers
Jul 10, 2018

Answer:

#x_1=1# or #x_2=-6#

Explanation:

Using the formula

#x_(1,2)=-p/2pmsqrt(p^2/4-q)#
#p=5,q=-6#

we get

#x_(1,2)=-5/2pmsqrt(25/4+6)#

#25/4+6=25/4+24/4=49/4#

so

#x_(1,2)-5/2pmsqrt(49/4)#
so we get

#x_1=1#

#x_2=-6#

Answer:

#x=-6, 1#

Explanation:

Given quadratic equation: #x^2+5x-6=0#

Comparing the above equation with the standard form of quadratic equation: #ax^2+bx+c=0# we get

#a=1, b=5# & #c=-6#

The roots of quadratic equation are given as follows

#x_{1, 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#

#=\frac{-5\pm\sqrt{5^2-4(1)(-6)}}{2(1)}#

#=\frac{-5\pm7}{2}#

#=-6, 1#

Jul 10, 2018

Answer:

#x=1,-6#

Explanation:

#x^2+5x-6=0# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=1#, #b=5#, #c=-6#

Quadratic formula

Substitute #0# for #y# and solve for #x#.

#x=(-b+-sqrt(b^2-4ac))/(2*a)#

Plug in the known values.

#x=(-5+-sqrt(49))/2#

Simplify.

#x=(-5+-7)/2#

#x=(-5+7)/2#, #(-5-7)/2#

Simplify.

#x=2/2#, #x=-12/2#

Simplify.

#x=1,-6#