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3x^2+5x-20=16+x-x2
We move all terms to the left:
3x^2+5x-20-(16+x-x2)=0
We add all the numbers together, and all the variables
3x^2-(+x-1x^2+16)+5x-20=0
We get rid of parentheses
3x^2+1x^2-x+5x-16-20=0
We add all the numbers together, and all the variables
4x^2+4x-36=0
a = 4; b = 4; c = -36;
Δ = b2-4ac
Δ = 42-4·4·(-36)
Δ = 592
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{37}}{2*4}=\frac{-4-4\sqrt{37}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{37}}{2*4}=\frac{-4+4\sqrt{37}}{8} $
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