Answer :
Solve for x: x^2+4x-5=16x
first take 16x from both sides of the equation
x²-12x-5=0
now we have a quadratic. we can solve it by using the quadratic formula
x=-b plus or minus the square root of b²-4ac all divided by 2a
where a=1, b=-12 and c=-5
plug these numbers into the formula
x=12 plus or minus the square root of 144-4x1x-5 all divided by 2
x=12 plus or minus the square root of 164 all divided by 2
or in decimal form x = {12.403124237, -0.403124237}
first take 16x from both sides of the equation
x²-12x-5=0
now we have a quadratic. we can solve it by using the quadratic formula
x=-b plus or minus the square root of b²-4ac all divided by 2a
where a=1, b=-12 and c=-5
plug these numbers into the formula
x=12 plus or minus the square root of 144-4x1x-5 all divided by 2
x=12 plus or minus the square root of 164 all divided by 2
or in decimal form x = {12.403124237, -0.403124237}
[tex]x^2+4x-5=16x\\
x^2-12x-5=0\\
x^2-12x+36-41=0\\
(x-6)^2=41\\
x-6=\sqrt{41} \vee x-6=-\sqrt{41}\\
x=6+\sqrt{41} \vee x=6-\sqrt{41}[/tex]