How do you solve #x^ { 2} + 4x = - 29#?

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Answer 1 · 2016-12-04T13:23:12

#x=-2+-5i#

#color(blue)(x^2+4x=-29#

Let's bring everything to the left hand side

Add #29# both sides

#rarrx^2+4x+29=0#

Now this is a Quadratic equation (In the form of #ax^2+bx+c=0#)

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

Where #a,b and c# are the coefficients of the terms

So #color(orange)(a=1,b=4 and c=29#

#rarrx=(-4+-sqrt(4^2-4(1)(29)))/(2(1))#

#rarrx=(-4+-sqrt(16-116))/(2)#

#rarrx=(-4+-sqrt(100*(-1)))/(2)#

#rarrx=(-4+-10sqrt(-1))/(2)#

#rarrx=cancel(-4)^-2/cancel2^1+-(cancel10^5sqrt(-1))/cancel(2)^1#

#color(green)(rArrx=-2+-5i#

Note: #(i=sqrt(-1))# it is an imaginary number

#:.color(green)(x=(-2+5i),(-2-5i)#