Question

what is the value of x in the equation
(4x+79)²+56+x√64=345+7x​

Answer 1

Solution :-

${(4x + 79)}^{2} + 56 + x \sqrt{64} = 345 + 7x$

${16x }^{2} + 6241 + 56 + 16x = 345 + 7x$

${16x}^{2} + 5952 + 16x = 7x$

${16x}^{2} + 16x - 7x + 5952 = 0$

${16x}^{2} + 9x + 5952 = 0$

Answer 2

Answer:

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