Question

4x^2+1204x_48528=0
solve the equation plzz.​

Answer 1

$\large\boxed{\fcolorbox{blue}{yellow} {Answer}}$

The roots of the given quadratic equation are

$x = - 137 \: \: or \: \: x = 36$

$\large\boxed{\fcolorbox{blue}{yellow} {Step - by - step explanation }}$

Given equation is

$4 {x}^{2} + 1204x - 48528 = 0$

Now,

$4 {x}^{2} + 1204x - 48528 = 0 \\ \\ {x}^{2} + 301x - 12132 = 0 \: \: \: ....(dividing \: by \: 4) \\ \\ {x}^{2} + 337x - 36x - 12132 = 0 \\ \\ x(x + 337) \: - 36(x + 337) \: = 0 \\ \\ (x + 337) \: (x - 36) \: = 0 \\ \\ now \\ (x + 137) = 0 \\ x + 137 = 0 \\ \\ x = - 137 \\ \\ or \\ \\ (x - 36) = 0 \\ \\ x = 36$

Hope it helps!

$<marquee> Mark as brainliest ✔️</marquee>$

Answer 2

WE HAVE,

$4x^{2} +1204x-48528 = 0\\\\(dividing by 4 both sides; we will get )\\\\x^{2} + 301x - 12132=0\\\\x^{2} +x(337-36)-12132=0\\\\x^{2} +337x - 36x - 12132 =0 \\\\x ( x + 337 ) -36 ( x + 337 ) =0 \\\\(x-36)(x+337)=0\\\\x = 36 or x = -337$

SO TWO ZEROES OF EQN ARE 36 AND -337.