How do you solve #7x - 10= - 6x ^ { 2}#?

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Answer 1 · 2017-09-22T01:28:43

Use the quadratic formula.

First, rearrange the equation to match the form #ax^2 + bx + c = 0#

#6x^2 + 7x - 10 = 0#

#a= 6#
#b = 7#
#c = -10#

Then using the quadratic formula: #x = (-b+-sqrt(b^2 - 4ac))/(2a)#
Substitute in the numbers:

#x = (-7+-sqrt(7^2 - 4(6*-10)))/(2*6)#

and simplify:

#x = (-7+-sqrt(289))/(12)#

#x = (-7+-17)/(12)#

So your two solutions would be:

#x = 5/6# and #x = -2#

Answer 2 · 2017-09-22T01:29:14

#7x-10=-6x^2#

#6x^2+7x-10=0#

#6x^2+12x-5x-10=0#

#6x*(x+2)-5*(x+2)=0#

#(6x-5)*(x+2)=0#

Hence #x_1=5/6# and #x_2=-2#

1) I transferred squared term to left side.

2) I decomposed polynomial using common multiplier.

3) I solved each multiplier.