.
#x^2+8x-25=0#
To solve for #x#, you use the quadratic formula which is:
for a quadratic equation of general form:
#ax^2+bx+c=0#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Here, #a=1#, #b=8#, and #c=-25#
Therefore,
#x=(-8+-sqrt(8^2-4(1)(-25)))/(2(1))=(-8+-sqrt(64+100))/2#
#x=(-8+-sqrt164)/2=(-8+-sqrt(4(41)))/2=(-8+-2sqrt41)/2#
#x=(cancelcolor(red)2(-4+-sqrt41))/cancelcolor(red)2=-4+-sqrt41#
Answers to other problems:
#A).#, #y=3(x-1)^2#
From: #(a-b)^2=a^2-2ab+b^2#, our #a=x#, and #b=1), #we have:
#y=3(x^2-2x+1)#
#y=3x^2-6x+3#
#B).#, #y=2(x-3)(x+4)#
#y=2(x^2+4x-3x-12)=2(x^2+x-12)#
#y=2x^2+2x-24#
#A).#, #y=6(x-7)^2-2#
This equation is in the vertex form:
#y=a(x-h)^2+k# where the coordinates of the vertex are #(h,k)#
Out #h=7#, and #k=-2#, therefore, vertex is #(7,-2)#
#B).# #y=-1/2(x+10)²-12#
Vertex is #(-10,-12)#
