How do you solve $(4x+5)^2=35x+29$ using the quadratic formula?

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Answer 1 · 2016-08-24T12:41:49

The Solns. are $x_1~=0.55425, x_2~=-1.80425$ .

Formula to find the roots $alpha, and beta$ of the quadr. eqn.

$ax^2+bx+c+0$ is, $alpha, beta = (-b+-sqrt(b^2-4ac))/(2a)$ .

Before proceed to solve the given eqn., let us first simplify it :

$(4x+5)^2=35x+29$ .

$rArr 16x^2+40x+25-35x-29=0, i.e., 4x^2+5x-4=0$ .

So, we have, $a=4, b=5, c=-4$

$:. alpha, beta =(-5+-sqrt(25+64))/8=(-5+-sqrt89)/8$

Taking, #sqrt89~=9.434, we have, the roots

$alpha~=0.55425, beta~=-1.80425$ .

Enjoy Maths.!

How do you solve (4x+5)^2=35x+29(4x+5)^2=35x+29 using the quadratic formula?