Ten times an integer is added to seven times it’s square. if the result is 152, what was the original number?

Required number is 4.

Step-by-step explanation:

Let the required number be a.

Given,

Sum of ten times the integer and seven times it’s square is 152.

= > Ten times of a + seven times of it's square = 152

= > 10( a ) + 7( a )^2 = 152

= > 10a + 7a^2 - 152 = 0

= > 7a^2 + 10a - 152 = 0

= > 7a^2 + ( 38 - 28 )a - 152 = 0

= > 7a^2 + 38a - 28a - 152 = 0

= > a( 7a + 38 ) - 4( 7a + 38 ) = 0

= > ( a - 4 )( 7a + 38 ) = 0

= > a = 4 or - 38 / 7

Hence the required number is 4.

a framing bias

step-by-step explanation:

rodrigo is subject to a framing bias i believe.

72 is the answer to this problem