5t2+10t-40=0

Factor 5 out of 5t2+10t-40.

Factor 5 out of 5t2.

5(t2)+10t-40=0

Factor 5 out of 10t.

5(t2)+5(2t)-40=0

Factor 5 out of -40.

5t2+5(2t)+5⋅-8=0

Factor 5 out of 5t2+5(2t).

5(t2+2t)+5⋅-8=0

Factor 5 out of 5(t2+2t)+5⋅-8.

5(t2+2t-8)=0

5(t2+2t-8)=0

Factor.

Factor t2+2t-8 using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -8 and whose sum is 2.

-2,4

Write the factored form using these integers.

5((t-2)(t+4))=0

5((t-2)(t+4))=0

Remove unnecessary parentheses.

5(t-2)(t+4)=0

5(t-2)(t+4)=0

5(t-2)(t+4)=0

Divide each term in 5(t-2)(t+4)=0 by 5.

5(t-2)(t+4)5=05

Simplify 5(t-2)(t+4)5.

Cancel the common factor of 5.

Cancel the common factor.

5(t-2)(t+4)5=05

Divide (t-2)(t+4) by 1.

(t-2)(t+4)=05

(t-2)(t+4)=05

Expand (t-2)(t+4) using the FOIL Method.

Apply the distributive property.

t(t+4)-2(t+4)=05

Apply the distributive property.

t⋅t+t⋅4-2(t+4)=05

Apply the distributive property.

t⋅t+t⋅4-2t-2⋅4=05

t⋅t+t⋅4-2t-2⋅4=05

Simplify and combine like terms.

Simplify each term.

Multiply t by t.

t2+t⋅4-2t-2⋅4=05

Move 4 to the left of t.

t2+4⋅t-2t-2⋅4=05

Multiply -2 by 4.

t2+4t-2t-8=05

t2+4t-2t-8=05

Subtract 2t from 4t.

t2+2t-8=05

t2+2t-8=05

t2+2t-8=05

Divide 0 by 5.

t2+2t-8=0

t2+2t-8=0

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -8 and whose sum is 2.

-2,4

Write the factored form using these integers.

(t-2)(t+4)=0

(t-2)(t+4)=0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

t-2=0

t+4=0

Set the first factor equal to 0.

t-2=0

Add 2 to both sides of the equation.

t=2

t=2

Set the next factor equal to 0.

t+4=0

Subtract 4 from both sides of the equation.

t=-4

t=-4

The final solution is all the values that make (t-2)(t+4)=0 true.

t=2,-4

Solve for t 5t^2+10t-40=0