Question

solve for (x+2)^2 and (x-5)^2 (x+2)^2= x^2+4x+4 (x-5)^2=x^2–10x+25 So solve the equation (x+2)^2=(x-5)^2+7 =>x^2+4x+4=x^2–10x+25+7 =>4x+4=32–10x =>4x+10x=32–4 =>14x=28 =>x=28/14 =>x=2 If you want to check if your answer is correct or not then you can put the value of x in the equation. So putting the

Answer 1

Explanation:

Quadratic Equation

Enter the coefficients for the Ax2 + Bx + C = 0 equation and Quadratic Equation will output the solutions (if they are not imaginary).

Quadratic Equation

Ax2 + Bx + C = 0

A =

B =

C =

X1 =

X2 =

If A=0, the equation is not quadratic.

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Answer 2

Answer:

solve for (x+2)^2 and (x-5)^2 (x+2)^2= x^2+4x+4 (x-5)^2=x^2–10x+25 So solve the equation (x+2)^2=(x-5)^2+7 =>x^2+4x+4=x^2–10x+25+7 =>4x+4=32–10x =>4x+10x=32–4 =>14x=28 =>x=28/14 =>x=2 If you want to check if your answer is correct or not then you can put the value of x in the equation. So putting the

Explanation:

solve for (x+2)^2 and (x-5)^2 (x+2)^2= x^2+4x+4 (x-5)^2=x^2–10x+25 So solve the equation (x+2)^2=(x-5)^2+7 =>x^2+4x+4=x^2–10x+25+7 =>4x+4=32–10x =>4x+10x=32–4 =>14x=28 =>x=28/14 =>x=2 If you want to check if your answer is correct or not then you can put the value of x in the equation. So putting the