Question 113570


{{{(9x+7)(7x+5)}}}

{{{(63x^2 + 45x +49x +35)}}}

{{{63x^2 + 94x +35}}}

We can solve it for {{{x}}} if we set this expression equal to {{{0}}}

{{{63x^2 + 94x +35 = 0}}}……..use square root formula


{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-94 +- sqrt (94^2 -4*63*35 )) / (2*63)}}}

{{{x[1,2]=(-94 +- sqrt (8836 - 8820)) / 126}}}


{{{x[1,2]=(-94 +- sqrt (16)) / 126}}}
	
{{{x[1,2]=(-94 +- 4) / 126}}}


{{{x[1]=(-94 + 4) / 126}}}

{{{x[1]= -90  / 126}}}
	
{{{x[1]= - 0.71}}}


{{{x[2]=(-94 - 4) / 126}}}

{{{x[2]= -98  / 126}}}

{{{x[2]= - 0.78}}}


Or, we can do it this way:

{{{(9x+7)(7x+5) = 0}}}….here we have a product of two binomials set to {{{0}}}; as we know, this  product will be equal to {{{0}}} if one binomial or both of them are equal to {{{0}}}.

{{{9x+7 = 0}}}….
{{{9x = -7}}}….
{{{x = -7/9}}}….

{{{x = - 0.78}}}………first solution


{{{7x+5 = 0}}}….
	
{{{7x = -5}}}….

{{{x = -5/7}}}….

{{{x = - 0.71}}}………..second solution