How do you solve $sqrt(3x+10)= 5-2x$ ?

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Answer 1 · 2016-04-11T20:25:50

$x_(1,2)={3/4,5}$

$sqrt(3x+10)=5-2x$

$(sqrt(3x+10))^2=(5-2x)^2$

$3x+10=(5-2x)^2$

$3x+10=25-20x+4x^2$

$4x^2-20x+25-3x-10=0$

$4x^2-23x+15=0$

$ax^2+bx+c=0$

$Delta=sqrt(b^2-4*a*c)$

$Delta=sqrt((-23)^2-4*4*15)$
$Delta=sqrt(529-240)$
$Delta=sqrt(289)$
$Delta=17$

$x_1=(-b-Delta)/(2*a)$

$x_1=(23-17)/(2*4)$

$x_1=6/8" "x_1=3/4$

$x_2=(-b+Delta)/(2*a)$

$x_2=(23+17)/(2*4)$

$x_2=20/8$

$x_2=5$

$x_(1,2)={3/4,5}$

How do you solve sqrt(3x+10)= 5-2xsqrt(3x+10)= 5-2x?