Determine which equation has the same solutions as the equation below. 4x^2+32x-28=0 A. (x+8)^2=57 B. 4(x+4)^2=36 C. 4(x+8)^2=28 D. (x+4)^2=23 D C ) A B

To determine which equation has the same solutions as the given equation $4x^{2}+32x-28=0$, we need to find the equation that can be transformed into the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. <br /><br />Let's analyze each option:<br /><br />A. $(x+8)^{2}=57$<br />Expanding this equation, we get $x^2 + 16x + 64 = 57$, which simplifies to $x^2 + 16x + 7 = 0$. This equation does not match the given equation.<br /><br />B. $4(x+4)^{2}=36anding this equation, we get $4(x^2 + 8x + 16) = 36$, which simplifies to $4x^2 + 32x + 64 = 36$. Subtracting 36 from both sides, we get $4x^2 + 32x + 28 = 0$. This equation matches the given equation.<br /><br />C. $4(x+8)^{2}=28$<br />Expanding this equation, we get $4(x^2 + 16x + 64) = 28$, which simplifies to $4x^2 + 64x + 256 = 28$. Subtracting 28 from both sides, we get $4x^2 + 64x + 228 = 0$. This equation does not match the given equation.<br /><br />D. $(x+4)^{2}=23$<br />Expanding this equation, we get $x^2 + 8x + 16 = 23$, which simplifies to $x^2 + 8x - 7 = 0$. This equation does not match the given equation.<br /><br />Therefore, the correct answer is B. $4(x+4)^{2}=36$.