5x^2+7x-4=12

Solution for 5x^2+7x-4=12 equation:

5x^2+7x-4=12

We move all terms to the left:

5x^2+7x-4-(12)=0

We add all the numbers together, and all the variables

5x^2+7x-16=0

a = 5; b = 7; c = -16;
Δ = b2-4ac
Δ = 72-4·5·(-16)
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3\sqrt{41}}{2*5}=\frac{-7-3\sqrt{41}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3\sqrt{41}}{2*5}=\frac{-7+3\sqrt{41}}{10}
$