3x^2+7x-6=-65/16

Solution for 3x^2+7x-6=-65/16 equation:

3x^2+7x-6=-65/16

We move all terms to the left:

3x^2+7x-6-(-65/16)=0

We get rid of parentheses

3x^2+7x-6+65/16=0

We multiply all the terms by the denominator


3x^2*16+7x*16+65-6*16=0

We add all the numbers together, and all the variables

3x^2*16+7x*16-31=0

Wy multiply elements

48x^2+112x-31=0

a = 48; b = 112; c = -31;
Δ = b2-4ac
Δ = 1122-4·48·(-31)
Δ = 18496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{18496}=136$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-136}{2*48}=\frac{-248}{96}
=-2+7/12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+136}{2*48}=\frac{24}{96}
=1/4
$