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√y +10 = y-2
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is y + 10 both inside the radical?
Yes, it is
√(y +10) = y-2 (square both side first, you will get) y+10 = y^2-4y+4 (put everything into the left side of the equation) -y^2-4y+y+10-4=0 (simplify) -y^2-3y+6 =0 (divide the equation by -1, just to make it easier) y^2+3y-6 =0 (you can't factor it, but you could use the square root formula) Square root formula is: - b ± √(b^2-4ac) a is a number which is times x^2 b is a number which is times x c is a constant (number which is not a variable or times a variable and it is not a zero) CAN you plug in the numbers into the simplified equation and solve, or still need explanations?
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