50625=B^2+225

asked by guest
on Mar 14, 2025 at 7:25 am



You asked:

Solve the equation 50625=B2+22550625 = {B}^{2} + 225 for the variable BB.

MathBot Answer:

B=6014224.49944B=6014224.49944\begin{aligned}B &= - 60 \sqrt{14} \approx -224.49944\\B &= 60 \sqrt{14} \approx 224.49944\end{aligned}

Solve by quadratic formula:

Subtract the right hand side from the left hand side of: 50625=B2+22550625 = B^{2} + 225 The result is a quadratic equation: 50400B2=050400 - B^{2} = 0

Given a quadratic equation ax2+bx+c=0a x^{2} + b x + c = 0, where aa, bb, cc are constants and a0a \ne 0, the solutions are given by the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} In this case a=1a=-1, b=0b=0, and c=50400c=50400.

The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:b24ac=02(4)50400=201600>0b^{2}-4ac = 0^{2} - \left(-4\right) 50400=201600 > 0 The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: B=(1)0+2016002(1)=6014224.49944B = \frac{\left(-1\right) 0 + \sqrt{201600}}{2 \left(-1\right)} = - 60 \sqrt{14} \approx -224.49944 B=(1)02016002(1)=6014224.49944B = \frac{\left(-1\right) 0 - \sqrt{201600}}{2 \left(-1\right)} = 60 \sqrt{14} \approx 224.49944