How do you solve #2x^{2}-28x+37=-111#?

2 Answers
Dec 18, 2017

No real solution. (i is an imaginary number)

Explanation:

#2x^2 -28x +37= -111#
Set all numbers to one side of the equation.
#2x^2 -28x +148= 0#
Divide by 2
#x^2 -14x +74= 0#
Use Quadratic Formula
# x=(\frac (-b\pm (\sqrt {b^{2)-4ac))){2a})#
Substitute
# x=(\frac (-(-14)\pm (\sqrt {(-14)^{2)-4(1)(74)))){2(1)})#
Solve
# x=(\frac (14\pm (\sqrt {(196)-296))){2(1)})#
# x=(\frac (14\pm (\sqrt {-100))){2(1)})#
# x=7\pm (10/2i)#
#x= 7+ 5i or x= 7- 5i#

Dec 18, 2017

#x=7+5i,##7-5i#

Explanation:

Solve:

#2x^2-28x+37=-111#

Add #111# to both sides of the equation.

#2x^2-28x+37+111=0#

Simplify.

#2x^2-28x+148=0# is a quadratic equation in standard form:

#ax^2+bx+c=0#,

where:

#a=2#, #b=-28#, and #c=148#

Solve for #x# using the quadratic formula.

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

Plug in the known values.

#x=(-(-28)+-sqrt(28^2-4*2*148))/(2*2)#

Simplify.

#x=(28+-sqrt(-400))/4#

Prime factorize #-400#.

#x=(28+-sqrt((2xx2)xx(2xx2)xx(5xx5)xx-1))/4#

Simplify.

#x=(28+-4xx5xxi)/4#

#x=(28+-20i)/4#

Factor out the common #4# in the numerator.

#x=(4(7+-5i))/4#

Divide #4# in the numerator by #4# in the denominator.

#x=7+-5i#

Solutions for #x#.

#x=7+5i,##7-5i#