Help me with these two equations
Solve for v
2v^2+4v-14=(v+1)^2
solve for y
y^2+6y+8=0
2 v^2 + 4 v - 14 = v^2 + 2 v + 1
v^2 + 2 v - 15 = 0
(v+5)(v-3) = 0
v = 3 or -5
the next one is easy
Ok Thank you
To solve the equation 2v^2 + 4v - 14 = (v + 1)^2 for v, you can follow these steps:
Step 1: Expand the squared term on the right side of the equation.
2v^2 + 4v - 14 = v^2 + 2v + 1
Step 2: Move all the terms to one side of the equation.
2v^2 + 4v - 14 - v^2 - 2v - 1 = 0
Simplifying the left side gives us:
v^2 + 2v - 15 = 0
Step 3: Factor the quadratic equation.
(v + 5)(v - 3) = 0
Step 4: Set each factor equal to zero and solve for v.
v + 5 = 0, v = -5
v - 3 = 0, v = 3
Therefore, the solutions for v are v = -5 and v = 3.
Now let's move on to solving the equation y^2 + 6y + 8 = 0 for y.
Step 1: Factor the quadratic equation.
(y + 2)(y + 4) = 0
Step 2: Set each factor equal to zero and solve for y.
y + 2 = 0, y = -2
y + 4 = 0, y = -4
Therefore, the solutions for y are y = -2 and y = -4.